Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A(c(b(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
A(c(b(b(x1)))) → B(c(x1))
B(b(b(x1))) → C(x1)
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(c(x1))) → A(b(b(x1)))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(x1))) → B(a(b(b(x1))))
B(b(c(x1))) → B(b(x1))
B(b(b(x1))) → C(b(a(c(x1))))
C(a(a(a(x1)))) → C(x1)
B(b(b(x1))) → B(a(c(x1)))
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A(c(b(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
A(c(b(b(x1)))) → B(c(x1))
B(b(b(x1))) → C(x1)
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(c(x1))) → A(b(b(x1)))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(x1))) → B(a(b(b(x1))))
B(b(c(x1))) → B(b(x1))
B(b(b(x1))) → C(b(a(c(x1))))
C(a(a(a(x1)))) → C(x1)
B(b(b(x1))) → B(a(c(x1)))
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(b(b(x1)))) → B(c(x1)) at position [0] we obtained the following new rules:
A(c(b(b(a(a(a(x0))))))) → B(a(c(x0)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A(c(b(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
A(c(b(b(a(a(a(x0))))))) → B(a(c(x0)))
B(b(c(x1))) → A(b(b(x1)))
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(b(x1))) → C(x1)
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(x1))) → B(b(x1))
B(b(c(x1))) → B(a(b(b(x1))))
B(b(b(x1))) → C(b(a(c(x1))))
C(a(a(a(x1)))) → C(x1)
B(b(b(x1))) → B(a(c(x1)))
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(c(x1))) → B(a(b(b(x1)))) at position [0] we obtained the following new rules:
B(b(c(c(x0)))) → B(a(b(a(b(b(x0))))))
B(b(c(b(b(x0))))) → B(a(b(c(b(a(c(x0)))))))
B(b(c(b(c(x0))))) → B(a(b(b(a(b(b(x0)))))))
B(b(c(b(x0)))) → B(a(c(b(a(c(x0))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A(c(b(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
A(c(b(b(a(a(a(x0))))))) → B(a(c(x0)))
B(b(c(b(c(x0))))) → B(a(b(b(a(b(b(x0)))))))
B(b(b(x1))) → C(x1)
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(c(x1))) → A(b(b(x1)))
B(b(c(c(x0)))) → B(a(b(a(b(b(x0))))))
B(b(c(b(b(x0))))) → B(a(b(c(b(a(c(x0)))))))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(b(x0)))) → B(a(c(b(a(c(x0))))))
B(b(c(x1))) → B(b(x1))
B(b(b(x1))) → C(b(a(c(x1))))
C(a(a(a(x1)))) → C(x1)
B(b(b(x1))) → B(a(c(x1)))
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(c(x1))) → A(b(b(x1))) at position [0] we obtained the following new rules:
B(b(c(b(x0)))) → A(c(b(a(c(x0)))))
B(b(c(b(b(x0))))) → A(b(c(b(a(c(x0))))))
B(b(c(c(x0)))) → A(b(a(b(b(x0)))))
B(b(c(b(c(x0))))) → A(b(b(a(b(b(x0))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A(c(b(b(x1)))) → C(x1)
B(b(c(b(b(x0))))) → A(b(c(b(a(c(x0))))))
A(c(b(b(x1)))) → B(b(c(x1)))
A(c(b(b(a(a(a(x0))))))) → B(a(c(x0)))
B(b(c(b(c(x0))))) → B(a(b(b(a(b(b(x0)))))))
B(b(c(b(c(x0))))) → A(b(b(a(b(b(x0))))))
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(b(x1))) → C(x1)
B(b(c(c(x0)))) → B(a(b(a(b(b(x0))))))
B(b(c(b(b(x0))))) → B(a(b(c(b(a(c(x0)))))))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(b(x0)))) → B(a(c(b(a(c(x0))))))
B(b(c(b(x0)))) → A(c(b(a(c(x0)))))
B(b(c(x1))) → B(b(x1))
B(b(c(c(x0)))) → A(b(a(b(b(x0)))))
B(b(b(x1))) → C(b(a(c(x1))))
C(a(a(a(x1)))) → C(x1)
B(b(b(x1))) → B(a(c(x1)))
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(b(x1))) → C(b(a(c(x1)))) at position [0] we obtained the following new rules:
B(b(b(a(a(a(x0)))))) → C(b(a(a(c(x0)))))
B(b(b(b(b(x0))))) → C(b(b(b(b(c(x0))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A(c(b(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
B(b(c(b(b(x0))))) → A(b(c(b(a(c(x0))))))
A(c(b(b(a(a(a(x0))))))) → B(a(c(x0)))
B(b(c(b(c(x0))))) → B(a(b(b(a(b(b(x0)))))))
B(b(b(x1))) → C(x1)
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(c(b(c(x0))))) → A(b(b(a(b(b(x0))))))
B(b(c(c(x0)))) → B(a(b(a(b(b(x0))))))
B(b(c(b(b(x0))))) → B(a(b(c(b(a(c(x0)))))))
B(b(b(a(a(a(x0)))))) → C(b(a(a(c(x0)))))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(b(b(b(x0))))) → C(b(b(b(b(c(x0))))))
B(b(c(b(x0)))) → B(a(c(b(a(c(x0))))))
B(b(c(x1))) → B(b(x1))
B(b(c(b(x0)))) → A(c(b(a(c(x0)))))
B(b(c(c(x0)))) → A(b(a(b(b(x0)))))
C(a(a(a(x1)))) → C(x1)
B(b(b(x1))) → B(a(c(x1)))
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(b(x1))) → B(a(c(x1))) at position [0] we obtained the following new rules:
B(b(b(b(b(x0))))) → B(b(b(b(c(x0)))))
B(b(b(a(a(a(x0)))))) → B(a(a(c(x0))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
Q DP problem:
The TRS P consists of the following rules:
A(c(b(b(x1)))) → C(x1)
B(b(c(b(b(x0))))) → A(b(c(b(a(c(x0))))))
A(c(b(b(x1)))) → B(b(c(x1)))
B(b(b(a(a(a(x0)))))) → B(a(a(c(x0))))
A(c(b(b(a(a(a(x0))))))) → B(a(c(x0)))
B(b(c(b(c(x0))))) → B(a(b(b(a(b(b(x0)))))))
B(b(c(b(c(x0))))) → A(b(b(a(b(b(x0))))))
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(b(x1))) → C(x1)
B(b(c(c(x0)))) → B(a(b(a(b(b(x0))))))
B(b(c(b(b(x0))))) → B(a(b(c(b(a(c(x0)))))))
B(b(b(b(b(x0))))) → B(b(b(b(c(x0)))))
B(b(b(a(a(a(x0)))))) → C(b(a(a(c(x0)))))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(b(b(b(x0))))) → C(b(b(b(b(c(x0))))))
B(b(c(b(x0)))) → B(a(c(b(a(c(x0))))))
B(b(c(b(x0)))) → A(c(b(a(c(x0)))))
B(b(c(x1))) → B(b(x1))
B(b(c(c(x0)))) → A(b(a(b(b(x0)))))
C(a(a(a(x1)))) → C(x1)
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.C: 0
c: x0
B: 0
a: 1
A: 0
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
B.0(b.0(c.0(b.1(x0)))) → B.0(a.0(c.0(b.1(a.1(c.1(x0))))))
A.0(c.0(b.0(b.1(x1)))) → C.0(x1)
B.0(b.0(b.0(x1))) → C.0(x1)
B.0(b.0(c.0(c.0(x0)))) → A.0(b.1(a.0(b.0(b.0(x0)))))
B.0(b.0(b.1(a.1(a.1(a.0(x0)))))) → B.1(a.1(a.0(c.0(x0))))
B.0(b.0(b.1(x1))) → A.0(c.1(x1))
C.1(a.1(a.1(a.1(x1)))) → C.1(x1)
B.0(b.0(b.0(b.0(b.0(x0))))) → B.0(b.0(b.0(b.0(c.0(x0)))))
A.0(c.0(b.0(b.1(x1)))) → B.0(b.0(b.1(c.1(x1))))
B.0(b.0(b.1(a.1(a.1(a.1(x0)))))) → B.1(a.1(a.1(c.1(x0))))
A.0(c.0(b.0(b.1(a.1(a.1(a.1(x0))))))) → B.0(a.1(c.1(x0)))
B.0(b.0(c.0(b.1(x0)))) → A.0(c.0(b.1(a.1(c.1(x0)))))
B.0(b.0(c.0(b.0(b.1(x0))))) → A.0(b.0(c.0(b.1(a.1(c.1(x0))))))
B.0(b.0(b.0(b.0(b.1(x0))))) → C.0(b.0(b.0(b.0(b.1(c.1(x0))))))
B.0(b.0(c.0(b.0(b.1(x0))))) → B.1(a.0(b.0(c.0(b.1(a.1(c.1(x0)))))))
B.0(b.0(c.0(c.0(x0)))) → B.0(a.0(b.1(a.0(b.0(b.0(x0))))))
B.0(b.0(b.1(a.1(a.1(a.0(x0)))))) → B.0(a.1(a.0(c.0(x0))))
B.0(b.0(c.0(b.1(x0)))) → B.1(a.0(c.0(b.1(a.1(c.1(x0))))))
C.1(a.1(a.1(a.1(x1)))) → C.0(x1)
B.0(b.0(c.0(b.0(b.1(x0))))) → B.0(a.0(b.0(c.0(b.1(a.1(c.1(x0)))))))
B.0(b.0(c.0(b.0(c.0(x0))))) → B.1(a.0(b.0(b.1(a.0(b.0(b.0(x0)))))))
B.0(b.0(b.1(x1))) → A.1(c.1(x1))
B.0(b.0(b.1(a.1(a.1(a.1(x0)))))) → C.0(b.1(a.1(a.1(c.1(x0)))))
A.0(c.0(b.0(b.0(x1)))) → B.0(b.0(b.0(c.0(x1))))
B.0(b.0(c.0(x1))) → B.0(b.0(x1))
B.0(b.0(c.0(b.0(b.0(x0))))) → B.1(a.0(b.0(c.0(b.1(a.0(c.0(x0)))))))
B.0(b.0(c.0(b.0(c.0(x0))))) → A.0(b.0(b.1(a.0(b.0(b.0(x0))))))
A.0(c.0(b.0(b.0(x1)))) → B.0(b.0(c.0(x1)))
B.0(b.1(c.1(x1))) → B.0(b.1(x1))
B.0(b.0(b.1(x1))) → C.0(x1)
B.0(b.0(c.0(b.0(x0)))) → A.0(c.0(b.1(a.0(c.0(x0)))))
C.1(a.1(a.1(a.0(x1)))) → C.0(x1)
B.0(b.0(b.1(a.1(a.1(a.0(x0)))))) → C.0(b.1(a.1(a.0(c.0(x0)))))
A.0(c.0(b.0(b.1(a.1(a.1(a.0(x0))))))) → B.0(a.0(c.0(x0)))
C.1(a.1(a.1(a.1(x1)))) → A.1(c.1(x1))
B.0(b.0(c.0(b.1(c.1(x0))))) → A.0(b.0(b.1(a.0(b.0(b.1(x0))))))
B.0(b.0(c.0(b.0(b.0(x0))))) → B.0(a.0(b.0(c.0(b.1(a.0(c.0(x0)))))))
B.0(b.0(b.1(x1))) → C.1(x1)
A.0(c.0(b.0(b.0(x1)))) → C.0(x1)
B.0(b.0(c.0(b.0(b.0(x0))))) → A.0(b.0(c.0(b.1(a.0(c.0(x0))))))
B.0(b.0(c.0(c.0(x0)))) → B.1(a.0(b.1(a.0(b.0(b.0(x0))))))
B.0(b.0(c.0(b.0(x0)))) → B.1(a.0(c.0(b.1(a.0(c.0(x0))))))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.0(b.0(b.0(b.1(c.1(x0)))))
A.0(c.0(b.0(b.1(a.1(a.1(a.0(x0))))))) → B.1(a.0(c.0(x0)))
C.1(a.1(a.1(a.1(x1)))) → A.0(c.1(x1))
B.0(b.0(c.0(x1))) → B.0(x1)
B.0(b.0(c.0(b.0(x0)))) → B.0(a.0(c.0(b.1(a.0(c.0(x0))))))
B.0(b.0(b.0(b.0(b.0(x0))))) → C.0(b.0(b.0(b.0(b.0(c.0(x0))))))
B.0(b.1(c.1(c.1(x0)))) → A.0(b.1(a.0(b.0(b.1(x0)))))
B.0(b.1(c.1(c.1(x0)))) → B.0(a.0(b.1(a.0(b.0(b.1(x0))))))
B.0(b.0(c.0(b.0(c.0(x0))))) → B.0(a.0(b.0(b.1(a.0(b.0(b.0(x0)))))))
B.0(b.0(c.0(b.1(c.1(x0))))) → B.1(a.0(b.0(b.1(a.0(b.0(b.1(x0)))))))
B.0(b.1(c.1(x1))) → B.1(x1)
B.0(b.0(c.0(b.1(c.1(x0))))) → B.0(a.0(b.0(b.1(a.0(b.0(b.1(x0)))))))
A.0(c.0(b.0(b.1(a.1(a.1(a.1(x0))))))) → B.1(a.1(c.1(x0)))
B.0(b.0(b.1(a.1(a.1(a.1(x0)))))) → B.0(a.1(a.1(c.1(x0))))
C.1(a.1(a.1(a.0(x1)))) → A.0(c.0(x1))
B.0(b.0(b.0(x1))) → A.0(c.0(x1))
B.0(b.1(c.1(x1))) → B.0(x1)
A.0(c.0(b.0(b.1(x1)))) → B.0(b.1(c.1(x1)))
A.0(c.0(b.0(b.1(x1)))) → C.1(x1)
B.0(b.1(c.1(c.1(x0)))) → B.1(a.0(b.1(a.0(b.0(b.1(x0))))))
The TRS R consists of the following rules:
c.1(x0) → c.0(x0)
b.0(b.0(c.0(x1))) → b.1(a.0(b.0(b.0(x1))))
b.0(b.1(c.1(x1))) → b.1(a.0(b.0(b.1(x1))))
b.0(b.0(b.1(x1))) → c.0(b.1(a.1(c.1(x1))))
b.1(x0) → b.0(x0)
a.0(c.0(b.0(b.1(x1)))) → b.0(b.0(b.1(c.1(x1))))
c.1(a.1(a.1(a.0(x1)))) → a.0(c.0(x1))
b.0(b.0(b.0(x1))) → c.0(b.1(a.0(c.0(x1))))
a.1(x0) → a.0(x0)
a.0(c.0(b.0(b.0(x1)))) → b.0(b.0(b.0(c.0(x1))))
c.1(a.1(a.1(a.1(x1)))) → a.1(c.1(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ SemLabProof2
Q DP problem:
The TRS P consists of the following rules:
B.0(b.0(c.0(b.1(x0)))) → B.0(a.0(c.0(b.1(a.1(c.1(x0))))))
A.0(c.0(b.0(b.1(x1)))) → C.0(x1)
B.0(b.0(b.0(x1))) → C.0(x1)
B.0(b.0(c.0(c.0(x0)))) → A.0(b.1(a.0(b.0(b.0(x0)))))
B.0(b.0(b.1(a.1(a.1(a.0(x0)))))) → B.1(a.1(a.0(c.0(x0))))
B.0(b.0(b.1(x1))) → A.0(c.1(x1))
C.1(a.1(a.1(a.1(x1)))) → C.1(x1)
B.0(b.0(b.0(b.0(b.0(x0))))) → B.0(b.0(b.0(b.0(c.0(x0)))))
A.0(c.0(b.0(b.1(x1)))) → B.0(b.0(b.1(c.1(x1))))
B.0(b.0(b.1(a.1(a.1(a.1(x0)))))) → B.1(a.1(a.1(c.1(x0))))
A.0(c.0(b.0(b.1(a.1(a.1(a.1(x0))))))) → B.0(a.1(c.1(x0)))
B.0(b.0(c.0(b.1(x0)))) → A.0(c.0(b.1(a.1(c.1(x0)))))
B.0(b.0(c.0(b.0(b.1(x0))))) → A.0(b.0(c.0(b.1(a.1(c.1(x0))))))
B.0(b.0(b.0(b.0(b.1(x0))))) → C.0(b.0(b.0(b.0(b.1(c.1(x0))))))
B.0(b.0(c.0(b.0(b.1(x0))))) → B.1(a.0(b.0(c.0(b.1(a.1(c.1(x0)))))))
B.0(b.0(c.0(c.0(x0)))) → B.0(a.0(b.1(a.0(b.0(b.0(x0))))))
B.0(b.0(b.1(a.1(a.1(a.0(x0)))))) → B.0(a.1(a.0(c.0(x0))))
B.0(b.0(c.0(b.1(x0)))) → B.1(a.0(c.0(b.1(a.1(c.1(x0))))))
C.1(a.1(a.1(a.1(x1)))) → C.0(x1)
B.0(b.0(c.0(b.0(b.1(x0))))) → B.0(a.0(b.0(c.0(b.1(a.1(c.1(x0)))))))
B.0(b.0(c.0(b.0(c.0(x0))))) → B.1(a.0(b.0(b.1(a.0(b.0(b.0(x0)))))))
B.0(b.0(b.1(x1))) → A.1(c.1(x1))
B.0(b.0(b.1(a.1(a.1(a.1(x0)))))) → C.0(b.1(a.1(a.1(c.1(x0)))))
A.0(c.0(b.0(b.0(x1)))) → B.0(b.0(b.0(c.0(x1))))
B.0(b.0(c.0(x1))) → B.0(b.0(x1))
B.0(b.0(c.0(b.0(b.0(x0))))) → B.1(a.0(b.0(c.0(b.1(a.0(c.0(x0)))))))
B.0(b.0(c.0(b.0(c.0(x0))))) → A.0(b.0(b.1(a.0(b.0(b.0(x0))))))
A.0(c.0(b.0(b.0(x1)))) → B.0(b.0(c.0(x1)))
B.0(b.1(c.1(x1))) → B.0(b.1(x1))
B.0(b.0(b.1(x1))) → C.0(x1)
B.0(b.0(c.0(b.0(x0)))) → A.0(c.0(b.1(a.0(c.0(x0)))))
C.1(a.1(a.1(a.0(x1)))) → C.0(x1)
B.0(b.0(b.1(a.1(a.1(a.0(x0)))))) → C.0(b.1(a.1(a.0(c.0(x0)))))
A.0(c.0(b.0(b.1(a.1(a.1(a.0(x0))))))) → B.0(a.0(c.0(x0)))
C.1(a.1(a.1(a.1(x1)))) → A.1(c.1(x1))
B.0(b.0(c.0(b.1(c.1(x0))))) → A.0(b.0(b.1(a.0(b.0(b.1(x0))))))
B.0(b.0(c.0(b.0(b.0(x0))))) → B.0(a.0(b.0(c.0(b.1(a.0(c.0(x0)))))))
B.0(b.0(b.1(x1))) → C.1(x1)
A.0(c.0(b.0(b.0(x1)))) → C.0(x1)
B.0(b.0(c.0(b.0(b.0(x0))))) → A.0(b.0(c.0(b.1(a.0(c.0(x0))))))
B.0(b.0(c.0(c.0(x0)))) → B.1(a.0(b.1(a.0(b.0(b.0(x0))))))
B.0(b.0(c.0(b.0(x0)))) → B.1(a.0(c.0(b.1(a.0(c.0(x0))))))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.0(b.0(b.0(b.1(c.1(x0)))))
A.0(c.0(b.0(b.1(a.1(a.1(a.0(x0))))))) → B.1(a.0(c.0(x0)))
C.1(a.1(a.1(a.1(x1)))) → A.0(c.1(x1))
B.0(b.0(c.0(x1))) → B.0(x1)
B.0(b.0(c.0(b.0(x0)))) → B.0(a.0(c.0(b.1(a.0(c.0(x0))))))
B.0(b.0(b.0(b.0(b.0(x0))))) → C.0(b.0(b.0(b.0(b.0(c.0(x0))))))
B.0(b.1(c.1(c.1(x0)))) → A.0(b.1(a.0(b.0(b.1(x0)))))
B.0(b.1(c.1(c.1(x0)))) → B.0(a.0(b.1(a.0(b.0(b.1(x0))))))
B.0(b.0(c.0(b.0(c.0(x0))))) → B.0(a.0(b.0(b.1(a.0(b.0(b.0(x0)))))))
B.0(b.0(c.0(b.1(c.1(x0))))) → B.1(a.0(b.0(b.1(a.0(b.0(b.1(x0)))))))
B.0(b.1(c.1(x1))) → B.1(x1)
B.0(b.0(c.0(b.1(c.1(x0))))) → B.0(a.0(b.0(b.1(a.0(b.0(b.1(x0)))))))
A.0(c.0(b.0(b.1(a.1(a.1(a.1(x0))))))) → B.1(a.1(c.1(x0)))
B.0(b.0(b.1(a.1(a.1(a.1(x0)))))) → B.0(a.1(a.1(c.1(x0))))
C.1(a.1(a.1(a.0(x1)))) → A.0(c.0(x1))
B.0(b.0(b.0(x1))) → A.0(c.0(x1))
B.0(b.1(c.1(x1))) → B.0(x1)
A.0(c.0(b.0(b.1(x1)))) → B.0(b.1(c.1(x1)))
A.0(c.0(b.0(b.1(x1)))) → C.1(x1)
B.0(b.1(c.1(c.1(x0)))) → B.1(a.0(b.1(a.0(b.0(b.1(x0))))))
The TRS R consists of the following rules:
c.1(x0) → c.0(x0)
b.0(b.0(c.0(x1))) → b.1(a.0(b.0(b.0(x1))))
b.0(b.1(c.1(x1))) → b.1(a.0(b.0(b.1(x1))))
b.0(b.0(b.1(x1))) → c.0(b.1(a.1(c.1(x1))))
b.1(x0) → b.0(x0)
a.0(c.0(b.0(b.1(x1)))) → b.0(b.0(b.1(c.1(x1))))
c.1(a.1(a.1(a.0(x1)))) → a.0(c.0(x1))
b.0(b.0(b.0(x1))) → c.0(b.1(a.0(c.0(x1))))
a.1(x0) → a.0(x0)
a.0(c.0(b.0(b.0(x1)))) → b.0(b.0(b.0(c.0(x1))))
c.1(a.1(a.1(a.1(x1)))) → a.1(c.1(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 29 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof2
Q DP problem:
The TRS P consists of the following rules:
B.0(b.0(c.0(b.0(x0)))) → A.0(c.0(b.1(a.0(c.0(x0)))))
B.0(b.0(c.0(b.1(x0)))) → B.0(a.0(c.0(b.1(a.1(c.1(x0))))))
B.0(b.0(c.0(c.0(x0)))) → A.0(b.1(a.0(b.0(b.0(x0)))))
A.0(c.0(b.0(b.1(a.1(a.1(a.0(x0))))))) → B.0(a.0(c.0(x0)))
B.0(b.0(c.0(b.1(c.1(x0))))) → A.0(b.0(b.1(a.0(b.0(b.1(x0))))))
B.0(b.0(b.1(x1))) → C.1(x1)
B.0(b.0(b.1(x1))) → A.0(c.1(x1))
C.1(a.1(a.1(a.1(x1)))) → C.1(x1)
B.0(b.0(b.0(b.0(b.0(x0))))) → B.0(b.0(b.0(b.0(c.0(x0)))))
A.0(c.0(b.0(b.1(x1)))) → B.0(b.0(b.1(c.1(x1))))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.0(b.0(b.0(b.1(c.1(x0)))))
C.1(a.1(a.1(a.1(x1)))) → A.0(c.1(x1))
A.0(c.0(b.0(b.1(a.1(a.1(a.1(x0))))))) → B.0(a.1(c.1(x0)))
B.0(b.0(c.0(x1))) → B.0(x1)
B.0(b.0(c.0(b.1(x0)))) → A.0(c.0(b.1(a.1(c.1(x0)))))
B.0(b.0(c.0(b.0(x0)))) → B.0(a.0(c.0(b.1(a.0(c.0(x0))))))
B.0(b.0(c.0(c.0(x0)))) → B.0(a.0(b.1(a.0(b.0(b.0(x0))))))
B.0(b.1(c.1(c.1(x0)))) → A.0(b.1(a.0(b.0(b.1(x0)))))
B.0(b.0(b.1(a.1(a.1(a.0(x0)))))) → B.0(a.1(a.0(c.0(x0))))
B.0(b.1(c.1(c.1(x0)))) → B.0(a.0(b.1(a.0(b.0(b.1(x0))))))
B.0(b.0(c.0(b.0(c.0(x0))))) → B.0(a.0(b.0(b.1(a.0(b.0(b.0(x0)))))))
B.0(b.0(c.0(b.1(c.1(x0))))) → B.0(a.0(b.0(b.1(a.0(b.0(b.1(x0)))))))
A.0(c.0(b.0(b.0(x1)))) → B.0(b.0(b.0(c.0(x1))))
B.0(b.0(c.0(x1))) → B.0(b.0(x1))
B.0(b.0(b.1(a.1(a.1(a.1(x0)))))) → B.0(a.1(a.1(c.1(x0))))
B.0(b.0(c.0(b.0(c.0(x0))))) → A.0(b.0(b.1(a.0(b.0(b.0(x0))))))
A.0(c.0(b.0(b.0(x1)))) → B.0(b.0(c.0(x1)))
C.1(a.1(a.1(a.0(x1)))) → A.0(c.0(x1))
B.0(b.0(b.0(x1))) → A.0(c.0(x1))
B.0(b.1(c.1(x1))) → B.0(x1)
B.0(b.1(c.1(x1))) → B.0(b.1(x1))
A.0(c.0(b.0(b.1(x1)))) → B.0(b.1(c.1(x1)))
A.0(c.0(b.0(b.1(x1)))) → C.1(x1)
The TRS R consists of the following rules:
c.1(x0) → c.0(x0)
b.0(b.0(c.0(x1))) → b.1(a.0(b.0(b.0(x1))))
b.0(b.1(c.1(x1))) → b.1(a.0(b.0(b.1(x1))))
b.0(b.0(b.1(x1))) → c.0(b.1(a.1(c.1(x1))))
b.1(x0) → b.0(x0)
a.0(c.0(b.0(b.1(x1)))) → b.0(b.0(b.1(c.1(x1))))
c.1(a.1(a.1(a.0(x1)))) → a.0(c.0(x1))
b.0(b.0(b.0(x1))) → c.0(b.1(a.0(c.0(x1))))
a.1(x0) → a.0(x0)
a.0(c.0(b.0(b.0(x1)))) → b.0(b.0(b.0(c.0(x1))))
c.1(a.1(a.1(a.1(x1)))) → a.1(c.1(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used.
Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
Q DP problem:
The TRS P consists of the following rules:
A(c(b(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
B(b(b(a(a(a(x0)))))) → B(a(a(c(x0))))
A(c(b(b(a(a(a(x0))))))) → B(a(c(x0)))
B(b(c(b(c(x0))))) → B(a(b(b(a(b(b(x0)))))))
B(b(c(b(c(x0))))) → A(b(b(a(b(b(x0))))))
B(b(b(x1))) → C(x1)
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(c(c(x0)))) → B(a(b(a(b(b(x0))))))
B(b(b(b(b(x0))))) → B(b(b(b(c(x0)))))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(b(x0)))) → B(a(c(b(a(c(x0))))))
B(b(c(x1))) → B(b(x1))
B(b(c(b(x0)))) → A(c(b(a(c(x0)))))
B(b(c(c(x0)))) → A(b(a(b(b(x0)))))
C(a(a(a(x1)))) → C(x1)
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
A(c(b(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
B(b(b(a(a(a(x0)))))) → B(a(a(c(x0))))
A(c(b(b(a(a(a(x0))))))) → B(a(c(x0)))
B(b(c(b(c(x0))))) → B(a(b(b(a(b(b(x0)))))))
B(b(c(b(c(x0))))) → A(b(b(a(b(b(x0))))))
B(b(b(x1))) → C(x1)
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(c(c(x0)))) → B(a(b(a(b(b(x0))))))
B(b(b(b(b(x0))))) → B(b(b(b(c(x0)))))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(b(x0)))) → B(a(c(b(a(c(x0))))))
B(b(c(x1))) → B(b(x1))
B(b(c(b(x0)))) → A(c(b(a(c(x0)))))
B(b(c(c(x0)))) → A(b(a(b(b(x0)))))
C(a(a(a(x1)))) → C(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
A(c(b(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
B(b(b(a(a(a(x0)))))) → B(a(a(c(x0))))
A(c(b(b(a(a(a(x0))))))) → B(a(c(x0)))
B(b(c(b(c(x0))))) → B(a(b(b(a(b(b(x0)))))))
B(b(c(b(c(x0))))) → A(b(b(a(b(b(x0))))))
B(b(b(x1))) → C(x1)
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(c(c(x0)))) → B(a(b(a(b(b(x0))))))
B(b(b(b(b(x0))))) → B(b(b(b(c(x0)))))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(b(x0)))) → B(a(c(b(a(c(x0))))))
B(b(c(x1))) → B(b(x1))
B(b(c(b(x0)))) → A(c(b(a(c(x0)))))
B(b(c(c(x0)))) → A(b(a(b(b(x0)))))
C(a(a(a(x1)))) → C(x1)
The set Q is empty.
We have obtained the following QTRS:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(c(b(b(x)))) → b(b(b(c(x))))
b(b(c(x))) → b(a(b(b(x))))
c(a(a(a(x)))) → a(c(x))
b(b(b(x))) → c(b(a(c(x))))
A(c(b(b(x)))) → C(x)
A(c(b(b(x)))) → B(b(c(x)))
B(b(b(a(a(a(x)))))) → B(a(a(c(x))))
A(c(b(b(a(a(a(x))))))) → B(a(c(x)))
B(b(c(b(c(x))))) → B(a(b(b(a(b(b(x)))))))
B(b(c(b(c(x))))) → A(b(b(a(b(b(x))))))
B(b(b(x))) → C(x)
A(c(b(b(x)))) → B(b(b(c(x))))
B(b(c(c(x)))) → B(a(b(a(b(b(x))))))
B(b(b(b(b(x))))) → B(b(b(b(c(x)))))
B(b(b(x))) → A(c(x))
C(a(a(a(x)))) → A(c(x))
B(b(c(x))) → B(x)
B(b(c(b(x)))) → B(a(c(b(a(c(x))))))
B(b(c(x))) → B(b(x))
B(b(c(b(x)))) → A(c(b(a(c(x)))))
B(b(c(c(x)))) → A(b(a(b(b(x)))))
C(a(a(a(x)))) → C(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(c(b(b(x)))) → b(b(b(c(x))))
b(b(c(x))) → b(a(b(b(x))))
c(a(a(a(x)))) → a(c(x))
b(b(b(x))) → c(b(a(c(x))))
A(c(b(b(x)))) → C(x)
A(c(b(b(x)))) → B(b(c(x)))
B(b(b(a(a(a(x)))))) → B(a(a(c(x))))
A(c(b(b(a(a(a(x))))))) → B(a(c(x)))
B(b(c(b(c(x))))) → B(a(b(b(a(b(b(x)))))))
B(b(c(b(c(x))))) → A(b(b(a(b(b(x))))))
B(b(b(x))) → C(x)
A(c(b(b(x)))) → B(b(b(c(x))))
B(b(c(c(x)))) → B(a(b(a(b(b(x))))))
B(b(b(b(b(x))))) → B(b(b(b(c(x)))))
B(b(b(x))) → A(c(x))
C(a(a(a(x)))) → A(c(x))
B(b(c(x))) → B(x)
B(b(c(b(x)))) → B(a(c(b(a(c(x))))))
B(b(c(x))) → B(b(x))
B(b(c(b(x)))) → A(c(b(a(c(x)))))
B(b(c(c(x)))) → A(b(a(b(b(x)))))
C(a(a(a(x)))) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(c(b(b(x)))) → b(b(b(c(x))))
b(b(c(x))) → b(a(b(b(x))))
c(a(a(a(x)))) → a(c(x))
b(b(b(x))) → c(b(a(c(x))))
A(c(b(b(x)))) → C(x)
A(c(b(b(x)))) → B(b(c(x)))
B(b(b(a(a(a(x)))))) → B(a(a(c(x))))
A(c(b(b(a(a(a(x))))))) → B(a(c(x)))
B(b(c(b(c(x))))) → B(a(b(b(a(b(b(x)))))))
B(b(c(b(c(x))))) → A(b(b(a(b(b(x))))))
B(b(b(x))) → C(x)
A(c(b(b(x)))) → B(b(b(c(x))))
B(b(c(c(x)))) → B(a(b(a(b(b(x))))))
B(b(b(b(b(x))))) → B(b(b(b(c(x)))))
B(b(b(x))) → A(c(x))
C(a(a(a(x)))) → A(c(x))
B(b(c(x))) → B(x)
B(b(c(b(x)))) → B(a(c(b(a(c(x))))))
B(b(c(x))) → B(b(x))
B(b(c(b(x)))) → A(c(b(a(c(x)))))
B(b(c(c(x)))) → A(b(a(b(b(x)))))
C(a(a(a(x)))) → C(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(c(b(b(x)))) → b(b(b(c(x))))
b(b(c(x))) → b(a(b(b(x))))
c(a(a(a(x)))) → a(c(x))
b(b(b(x))) → c(b(a(c(x))))
A(c(b(b(x)))) → C(x)
A(c(b(b(x)))) → B(b(c(x)))
B(b(b(a(a(a(x)))))) → B(a(a(c(x))))
A(c(b(b(a(a(a(x))))))) → B(a(c(x)))
B(b(c(b(c(x))))) → B(a(b(b(a(b(b(x)))))))
B(b(c(b(c(x))))) → A(b(b(a(b(b(x))))))
B(b(b(x))) → C(x)
A(c(b(b(x)))) → B(b(b(c(x))))
B(b(c(c(x)))) → B(a(b(a(b(b(x))))))
B(b(b(b(b(x))))) → B(b(b(b(c(x)))))
B(b(b(x))) → A(c(x))
C(a(a(a(x)))) → A(c(x))
B(b(c(x))) → B(x)
B(b(c(b(x)))) → B(a(c(b(a(c(x))))))
B(b(c(x))) → B(b(x))
B(b(c(b(x)))) → A(c(b(a(c(x)))))
B(b(c(c(x)))) → A(b(a(b(b(x)))))
C(a(a(a(x)))) → C(x)
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
B1(c(b(B(x)))) → B1(c(a(B(x))))
B1(b(c(A(x)))) → B1(B(x))
B1(b(b(x))) → A1(b(c(x)))
C1(b(b(x))) → B1(a(b(x)))
C1(b(b(x))) → A1(b(x))
C1(c(b(B(x)))) → B1(A(x))
C1(b(c(b(B(x))))) → A1(b(b(A(x))))
C1(c(b(B(x)))) → B1(b(a(b(a(B(x))))))
C1(b(c(b(B(x))))) → B1(b(A(x)))
C1(b(c(b(B(x))))) → B1(b(a(b(b(a(B(x)))))))
B1(b(B(x))) → C1(A(x))
C1(b(c(b(B(x))))) → B1(b(a(B(x))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(x))) → C1(x)
B1(c(b(B(x)))) → C1(a(b(c(A(x)))))
C1(b(b(x))) → B1(b(a(b(x))))
C1(b(c(b(B(x))))) → B1(A(x))
B1(c(b(B(x)))) → C1(A(x))
B1(b(c(A(x)))) → B1(b(B(x)))
B1(c(b(B(x)))) → B1(c(A(x)))
B1(b(c(a(x)))) → B1(b(x))
A1(a(a(c(x)))) → A1(x)
B1(c(b(B(x)))) → A1(B(x))
B1(b(c(a(x)))) → C1(b(b(b(x))))
C1(b(c(b(B(x))))) → B1(b(a(b(b(A(x))))))
A1(a(a(C(x)))) → C1(A(x))
C1(b(c(b(B(x))))) → B1(a(b(b(a(B(x))))))
A1(a(a(b(b(B(x)))))) → C1(a(a(B(x))))
C1(c(b(B(x)))) → B1(a(b(A(x))))
B1(b(c(A(x)))) → C1(b(B(x)))
B1(c(b(B(x)))) → C1(a(b(c(a(B(x))))))
B1(b(c(A(x)))) → C1(b(b(B(x))))
C1(c(b(B(x)))) → A1(B(x))
B1(b(c(a(x)))) → B1(x)
C1(c(b(B(x)))) → B1(a(b(a(B(x)))))
A1(a(a(b(b(c(A(x))))))) → C1(a(B(x)))
C1(b(c(b(B(x))))) → A1(B(x))
C1(b(c(b(B(x))))) → B1(a(B(x)))
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
A1(a(a(c(x)))) → C1(a(x))
B1(c(b(B(x)))) → C1(a(B(x)))
B1(c(b(B(x)))) → A1(b(c(A(x))))
C1(c(b(B(x)))) → A1(b(A(x)))
B1(b(b(x))) → B1(c(x))
A1(a(a(b(b(B(x)))))) → A1(B(x))
A1(a(a(b(b(c(A(x))))))) → A1(B(x))
C1(b(c(b(B(x))))) → B1(a(b(b(A(x)))))
C1(c(b(B(x)))) → A1(b(a(B(x))))
B1(b(b(x))) → C1(a(b(c(x))))
C1(c(b(B(x)))) → B1(a(B(x)))
C1(c(b(B(x)))) → B1(b(a(b(A(x)))))
A1(a(a(b(b(B(x)))))) → A1(a(B(x)))
C1(b(c(b(B(x))))) → A1(b(b(a(B(x)))))
B1(c(b(B(x)))) → A1(b(c(a(B(x)))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
B1(c(b(B(x)))) → B1(c(a(B(x))))
B1(b(c(A(x)))) → B1(B(x))
B1(b(b(x))) → A1(b(c(x)))
C1(b(b(x))) → B1(a(b(x)))
C1(b(b(x))) → A1(b(x))
C1(c(b(B(x)))) → B1(A(x))
C1(b(c(b(B(x))))) → A1(b(b(A(x))))
C1(c(b(B(x)))) → B1(b(a(b(a(B(x))))))
C1(b(c(b(B(x))))) → B1(b(A(x)))
C1(b(c(b(B(x))))) → B1(b(a(b(b(a(B(x)))))))
B1(b(B(x))) → C1(A(x))
C1(b(c(b(B(x))))) → B1(b(a(B(x))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(x))) → C1(x)
B1(c(b(B(x)))) → C1(a(b(c(A(x)))))
C1(b(b(x))) → B1(b(a(b(x))))
C1(b(c(b(B(x))))) → B1(A(x))
B1(c(b(B(x)))) → C1(A(x))
B1(b(c(A(x)))) → B1(b(B(x)))
B1(c(b(B(x)))) → B1(c(A(x)))
B1(b(c(a(x)))) → B1(b(x))
A1(a(a(c(x)))) → A1(x)
B1(c(b(B(x)))) → A1(B(x))
B1(b(c(a(x)))) → C1(b(b(b(x))))
C1(b(c(b(B(x))))) → B1(b(a(b(b(A(x))))))
A1(a(a(C(x)))) → C1(A(x))
C1(b(c(b(B(x))))) → B1(a(b(b(a(B(x))))))
A1(a(a(b(b(B(x)))))) → C1(a(a(B(x))))
C1(c(b(B(x)))) → B1(a(b(A(x))))
B1(b(c(A(x)))) → C1(b(B(x)))
B1(c(b(B(x)))) → C1(a(b(c(a(B(x))))))
B1(b(c(A(x)))) → C1(b(b(B(x))))
C1(c(b(B(x)))) → A1(B(x))
B1(b(c(a(x)))) → B1(x)
C1(c(b(B(x)))) → B1(a(b(a(B(x)))))
A1(a(a(b(b(c(A(x))))))) → C1(a(B(x)))
C1(b(c(b(B(x))))) → A1(B(x))
C1(b(c(b(B(x))))) → B1(a(B(x)))
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
A1(a(a(c(x)))) → C1(a(x))
B1(c(b(B(x)))) → C1(a(B(x)))
B1(c(b(B(x)))) → A1(b(c(A(x))))
C1(c(b(B(x)))) → A1(b(A(x)))
B1(b(b(x))) → B1(c(x))
A1(a(a(b(b(B(x)))))) → A1(B(x))
A1(a(a(b(b(c(A(x))))))) → A1(B(x))
C1(b(c(b(B(x))))) → B1(a(b(b(A(x)))))
C1(c(b(B(x)))) → A1(b(a(B(x))))
B1(b(b(x))) → C1(a(b(c(x))))
C1(c(b(B(x)))) → B1(a(B(x)))
C1(c(b(B(x)))) → B1(b(a(b(A(x)))))
A1(a(a(b(b(B(x)))))) → A1(a(B(x)))
C1(b(c(b(B(x))))) → A1(b(b(a(B(x)))))
B1(c(b(B(x)))) → A1(b(c(a(B(x)))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 42 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(x))) → B1(a(b(x)))
B1(b(b(x))) → C1(x)
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
C1(b(b(x))) → B1(b(a(b(x))))
B1(b(c(A(x)))) → C1(b(b(B(x))))
B1(b(b(x))) → B1(c(x))
B1(b(b(x))) → C1(a(b(c(x))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(b(b(x))) → B1(b(a(b(x)))) at position [0] we obtained the following new rules:
C1(b(b(b(c(A(x0)))))) → B1(b(a(C(x0))))
C1(b(b(b(B(x0))))) → B1(b(a(c(A(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(c(b(B(x0)))))) → B1(b(a(c(a(b(c(A(x0))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
C1(b(b(b(B(x0))))) → B1(b(a(C(x0))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(c(b(B(x0)))))) → B1(b(a(c(a(b(c(a(B(x0)))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C1(b(b(x))) → B1(a(b(x)))
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(b(B(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(B(x0))))) → B1(b(a(C(x0))))
B1(b(b(x))) → C1(a(b(c(x))))
C1(b(b(c(b(B(x0)))))) → B1(b(a(c(a(b(c(a(B(x0)))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(C(x0))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(B(x0))))) → B1(b(a(c(A(x0)))))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(c(A(x)))) → C1(b(b(B(x))))
B1(b(b(x))) → B1(c(x))
C1(b(b(c(b(B(x0)))))) → B1(b(a(c(a(b(c(A(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C1(b(b(x))) → B1(a(b(x)))
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(b(B(x0)))))))
B1(b(b(x))) → C1(a(b(c(x))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(x))) → B1(c(x))
B1(b(c(A(x)))) → C1(b(b(B(x))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(b(b(x))) → B1(a(b(x))) at position [0] we obtained the following new rules:
C1(b(b(b(B(x0))))) → B1(a(c(A(x0))))
C1(b(b(b(c(A(x0)))))) → B1(a(C(x0)))
C1(b(b(c(b(B(x0)))))) → B1(a(c(a(b(c(A(x0)))))))
C1(b(b(c(b(B(x0)))))) → B1(a(c(a(b(c(a(B(x0))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(B(x0))))) → B1(a(C(x0)))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(b(B(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(B(x0))))) → B1(a(C(x0)))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(b(x))) → B1(c(x))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(b(x))
C1(b(b(b(c(A(x0)))))) → B1(a(C(x0)))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(x))) → C1(a(b(c(x))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(B(x0))))) → B1(a(c(A(x0))))
C1(b(b(c(b(B(x0)))))) → B1(a(c(a(b(c(A(x0)))))))
C1(b(b(c(b(B(x0)))))) → B1(a(c(a(b(c(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(A(x)))) → C1(b(b(B(x))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(b(B(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
B1(b(b(x))) → C1(a(b(c(x))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(x))) → B1(c(x))
B1(b(c(A(x)))) → C1(b(b(B(x))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(b(B(x0))))))) at position [0] we obtained the following new rules:
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(C(x0)))))
C1(b(b(b(c(A(y0)))))) → B1(b(a(b(b(a(b(B(y0))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(c(A(x0))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(C(x0)))))
C1(b(b(b(c(A(y0)))))) → B1(b(a(b(b(a(b(B(y0))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(x))) → C1(a(b(c(x))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(c(A(x0))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(c(A(x)))) → C1(b(b(B(x))))
B1(b(b(x))) → B1(c(x))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
B1(b(b(x))) → C1(a(b(c(x))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(x))) → B1(c(x))
B1(b(c(A(x)))) → C1(b(b(B(x))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(x))) → C1(a(b(c(x)))) at position [0] we obtained the following new rules:
B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0)))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(b(b(B(x0))))) → C1(a(c(a(b(c(a(B(x0))))))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
B1(b(b(b(B(x0))))) → C1(a(c(a(b(c(A(x0)))))))
B1(b(b(b(B(x0))))) → C1(a(b(B(x0))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(b(b(B(x0))))) → C1(a(b(B(x0))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(x))) → B1(c(x))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(B(x0))))) → C1(a(c(a(b(c(a(B(x0))))))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(A(x)))) → C1(b(b(B(x))))
B1(b(b(b(B(x0))))) → C1(a(c(a(b(c(A(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(x))) → B1(c(x))
B1(b(c(A(x)))) → C1(b(b(B(x))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(x))) → B1(c(x)) at position [0] we obtained the following new rules:
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(b(b(B(x0))))) → B1(b(B(x0)))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
B1(b(b(b(B(x0))))) → B1(B(x0))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(B(x0))))) → B1(b(B(x0)))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0)))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(B(x0))))) → B1(B(x0))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(A(x)))) → C1(b(b(B(x))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0)))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(A(x)))) → C1(b(b(B(x))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(c(A(x)))) → C1(b(b(B(x)))) at position [0] we obtained the following new rules:
B1(b(c(A(x0)))) → C1(C(x0))
B1(b(c(A(x0)))) → C1(c(A(x0)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(A(x0)))) → C1(C(x0))
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
B1(b(c(A(x0)))) → C1(c(A(x0)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0)))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(B(x))))) → C1(b(b(b(B(x)))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0)))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(B(x))))) → C1(b(b(b(B(x))))) at position [0] we obtained the following new rules:
B1(b(b(b(B(x0))))) → C1(b(C(x0)))
B1(b(b(b(B(x0))))) → C1(b(c(A(x0))))
B1(b(b(b(B(y0))))) → C1(c(a(b(c(B(y0))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0)))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(B(x0))))) → C1(b(C(x0)))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(B(x0))))) → C1(b(c(A(x0))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(b(b(B(y0))))) → C1(c(a(b(c(B(y0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0)))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(B(x0))))) → C1(a(b(b(B(x0))))) at position [0] we obtained the following new rules:
B1(b(b(b(B(x0))))) → C1(a(c(A(x0))))
B1(b(b(b(B(x0))))) → C1(a(C(x0)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(B(x0))))) → C1(a(c(A(x0))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(B(x0))))) → C1(a(C(x0)))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(a(B(x0))))))))) at position [0] we obtained the following new rules:
B1(b(b(c(b(B(y0)))))) → C1(a(c(a(b(c(a(b(a(B(y0))))))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(y0)))))) → C1(a(c(a(b(c(a(b(a(B(y0))))))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0))))))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(a(B(x0)))))))))) at position [0] we obtained the following new rules:
B1(b(b(b(c(b(B(y0))))))) → C1(a(c(a(b(c(a(b(b(a(B(y0)))))))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(b(c(b(B(y0))))))) → C1(a(c(a(b(c(a(b(b(a(B(y0)))))))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(b(b(b(c(A(x0)))))) → B1(b(a(c(b(B(x0)))))) at position [0] we obtained the following new rules:
C1(b(b(b(c(A(x0)))))) → B1(b(a(b(B(x0)))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(B(x0))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(B(x0))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(b(a(b(B(x0)))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0))))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(c(b(B(x0)))))) → C1(a(b(b(b(a(b(A(x0)))))))) at position [0] we obtained the following new rules:
B1(b(b(c(b(B(y0)))))) → C1(a(c(a(b(c(a(b(A(y0)))))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(c(b(B(y0)))))) → C1(a(c(a(b(c(a(b(A(y0)))))))))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0)))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(c(b(B(x0))))))) → C1(a(b(b(b(a(b(b(A(x0))))))))) at position [0] we obtained the following new rules:
B1(b(b(b(c(b(B(y0))))))) → C1(a(c(a(b(c(a(b(b(A(y0))))))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
B1(b(b(b(c(b(B(y0))))))) → C1(a(c(a(b(c(a(b(b(A(y0))))))))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0)))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(b(b(b(c(A(x0)))))) → B1(a(c(b(B(x0))))) at position [0] we obtained the following new rules:
C1(b(b(b(c(A(x0)))))) → B1(a(B(x0)))
C1(b(b(b(c(A(x0)))))) → B1(a(b(B(x0))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C1(b(b(b(c(A(x0)))))) → B1(a(B(x0)))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(c(A(x0)))))) → B1(a(b(B(x0))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(b(b(b(c(A(x0)))))) → B1(a(c(b(b(B(x0)))))) at position [0] we obtained the following new rules:
C1(b(b(b(c(A(y0)))))) → B1(a(b(b(a(b(B(y0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(c(A(x0)))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(C(x0))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
C1(b(b(b(c(A(x0)))))) → B1(a(c(C(x0))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(c(A(y0)))))) → B1(a(b(b(a(b(B(y0)))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
C1(b(b(b(c(A(x0)))))) → B1(a(c(c(A(x0)))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
B1(b(b(b(b(x0))))) → C1(a(b(b(b(a(b(x0)))))))
C1(b(b(b(b(x0))))) → B1(a(c(a(b(c(x0))))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
C1(b(b(b(b(b(B(x0))))))) → B1(a(c(b(b(b(B(x0)))))))
C1(b(b(b(c(a(x0)))))) → B1(a(c(b(b(b(x0))))))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.C: 0
c: 0
B: 0
a: 1
A: 0
B1: 0
b: 0
C1: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.1(a.0(c.0(b.0(b.0(b.0(B.0(x0)))))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.0(a.0(c.0(b.0(b.0(b.0(x0))))))
B1.0(b.0(b.0(x))) → C1.0(x)
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.0(x0))))))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(x)
B1.0(b.0(c.1(a.1(x)))) → B1.0(b.0(b.1(x)))
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.0(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(x0)))))))
B1.0(b.0(b.0(b.0(b.0(x0))))) → C1.1(a.0(b.0(b.0(b.1(a.0(b.0(x0)))))))
C1.0(b.0(b.0(b.0(c.1(a.1(x0)))))) → B1.0(a.0(c.0(b.0(b.0(b.1(x0))))))
B1.0(b.0(c.1(a.0(x)))) → C1.0(b.0(b.0(b.0(x))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.0(x0))))))))
B1.0(b.0(b.0(b.0(b.1(x0))))) → C1.0(a.0(b.0(b.0(b.1(a.0(b.1(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.0(x0)))))))
B1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.0(b.1(a.0(b.1(x0)))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(b.0(x)))
B1.0(b.0(b.1(x))) → C1.1(x)
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.1(a.0(c.1(a.0(b.0(c.0(x0))))))
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.1(a.0(c.1(a.0(b.0(c.1(x0))))))
B1.0(b.0(c.1(a.1(x)))) → C1.0(b.0(b.0(b.1(x))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.0(x0)))))))
B1.0(b.0(c.1(a.1(x)))) → B1.1(x)
C1.0(b.0(b.0(b.0(c.1(a.1(x0)))))) → B1.1(a.0(c.0(b.0(b.0(b.1(x0))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.0(a.0(c.0(b.0(b.0(b.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.1(a.0(c.0(b.0(b.0(b.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.0(a.0(c.0(b.0(b.0(b.0(B.0(x0)))))))
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.1(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(a.0(c.1(a.0(b.0(c.0(x0))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.1(x0))))))
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(a.0(c.1(a.0(b.0(c.1(x0))))))
C1.0(b.0(b.0(b.0(c.1(a.1(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.1(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.0(x0))))))
B1.0(b.0(b.1(x))) → C1.0(x)
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(x))
B1.0(b.0(b.0(b.0(b.1(x0))))) → C1.1(a.0(b.0(b.0(b.1(a.0(b.1(x0)))))))
B1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.0(b.1(a.0(b.0(x0)))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.1(a.0(c.0(b.0(b.0(b.0(x0))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.1(x0)))))))
B1.0(b.0(b.0(b.0(b.0(x0))))) → C1.0(a.0(b.0(b.0(b.1(a.0(b.0(x0)))))))
B1.0(b.0(c.1(a.1(x)))) → B1.0(x)
B1.0(b.0(c.1(a.1(x)))) → B1.0(b.1(x))
The TRS R consists of the following rules:
b.0(b.0(b.0(b.0(B.1(x))))) → c.0(b.0(b.0(b.0(B.1(x)))))
b.0(b.0(B.0(x))) → c.0(A.0(x))
A.1(x0) → A.0(x0)
c.0(b.0(B.1(x))) → B.1(x)
a.1(x0) → a.0(x0)
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.0(A.0(x)))))
B.1(x0) → B.0(x0)
c.0(b.0(b.1(x))) → b.0(b.1(a.0(b.1(x))))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.1(a.0(B.0(x))))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.0(x)))))))
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(b.0(B.0(x))))
b.0(b.0(b.0(x))) → c.1(a.0(b.0(c.0(x))))
a.1(a.1(a.0(c.0(x)))) → c.1(a.0(x))
c.0(b.0(B.0(x))) → B.0(x)
b.0(b.0(c.1(a.0(x)))) → c.0(b.0(b.0(b.0(x))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(b.0(B.1(x))))
b.0(b.0(c.0(A.0(x)))) → C.0(x)
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(B.0(x)))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.1(x)))))))
c.0(b.0(B.0(x))) → b.0(B.0(x))
a.1(a.1(a.0(C.1(x)))) → C.1(x)
c.0(b.0(b.0(x))) → b.0(b.1(a.0(b.0(x))))
a.1(a.1(a.0(b.0(b.0(B.0(x)))))) → c.1(a.1(a.0(B.0(x))))
a.1(a.1(a.0(b.0(b.0(c.0(A.1(x))))))) → c.1(a.0(B.1(x)))
b.1(x0) → b.0(x0)
b.0(b.0(B.1(x))) → c.0(A.1(x))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.1(a.0(B.1(x))))))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.0(A.0(x)))))
a.1(a.1(a.0(C.0(x)))) → C.0(x)
b.0(b.0(b.0(b.0(B.0(x))))) → c.0(b.0(b.0(b.0(B.0(x)))))
b.0(b.0(c.1(a.1(x)))) → c.0(b.0(b.0(b.1(x))))
a.1(a.1(a.0(b.0(b.0(B.1(x)))))) → c.1(a.1(a.0(B.1(x))))
a.1(a.1(a.0(C.0(x)))) → c.0(A.0(x))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(B.1(x)))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.0(A.0(x))))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.1(a.0(B.1(x))))))
b.0(b.0(B.0(x))) → C.0(x)
c.0(b.0(B.1(x))) → b.0(B.1(x))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.0(A.1(x))))))
a.1(a.1(a.0(b.0(b.0(c.0(A.0(x))))))) → c.1(a.0(B.0(x)))
c.1(x0) → c.0(x0)
b.0(b.0(c.0(A.1(x)))) → C.1(x)
C.1(x0) → C.0(x0)
a.1(a.1(a.0(c.1(x)))) → c.1(a.1(x))
b.0(b.0(B.1(x))) → C.1(x)
a.1(a.1(a.0(C.1(x)))) → c.0(A.1(x))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.1(a.0(B.0(x))))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.0(A.1(x)))))
b.0(b.0(b.1(x))) → c.1(a.0(b.0(c.1(x))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.0(A.1(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ SemLabProof2
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.1(a.0(c.0(b.0(b.0(b.0(B.0(x0)))))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.0(a.0(c.0(b.0(b.0(b.0(x0))))))
B1.0(b.0(b.0(x))) → C1.0(x)
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.0(x0))))))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(x)
B1.0(b.0(c.1(a.1(x)))) → B1.0(b.0(b.1(x)))
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.0(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(x0)))))))
B1.0(b.0(b.0(b.0(b.0(x0))))) → C1.1(a.0(b.0(b.0(b.1(a.0(b.0(x0)))))))
C1.0(b.0(b.0(b.0(c.1(a.1(x0)))))) → B1.0(a.0(c.0(b.0(b.0(b.1(x0))))))
B1.0(b.0(c.1(a.0(x)))) → C1.0(b.0(b.0(b.0(x))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.0(x0))))))))
B1.0(b.0(b.0(b.0(b.1(x0))))) → C1.0(a.0(b.0(b.0(b.1(a.0(b.1(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.0(x0)))))))
B1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.0(b.1(a.0(b.1(x0)))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(b.0(x)))
B1.0(b.0(b.1(x))) → C1.1(x)
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.1(a.0(c.1(a.0(b.0(c.0(x0))))))
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.1(a.0(c.1(a.0(b.0(c.1(x0))))))
B1.0(b.0(c.1(a.1(x)))) → C1.0(b.0(b.0(b.1(x))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.0(x0)))))))
B1.0(b.0(c.1(a.1(x)))) → B1.1(x)
C1.0(b.0(b.0(b.0(c.1(a.1(x0)))))) → B1.1(a.0(c.0(b.0(b.0(b.1(x0))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.0(a.0(c.0(b.0(b.0(b.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.1(a.0(c.0(b.0(b.0(b.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.0(a.0(c.0(b.0(b.0(b.0(B.0(x0)))))))
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.1(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(a.0(c.1(a.0(b.0(c.0(x0))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.1(x0))))))
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(a.0(c.1(a.0(b.0(c.1(x0))))))
C1.0(b.0(b.0(b.0(c.1(a.1(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.1(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.0(x0))))))
B1.0(b.0(b.1(x))) → C1.0(x)
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(x))
B1.0(b.0(b.0(b.0(b.1(x0))))) → C1.1(a.0(b.0(b.0(b.1(a.0(b.1(x0)))))))
B1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.0(b.1(a.0(b.0(x0)))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.1(a.0(c.0(b.0(b.0(b.0(x0))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.1(x0)))))))
B1.0(b.0(b.0(b.0(b.0(x0))))) → C1.0(a.0(b.0(b.0(b.1(a.0(b.0(x0)))))))
B1.0(b.0(c.1(a.1(x)))) → B1.0(x)
B1.0(b.0(c.1(a.1(x)))) → B1.0(b.1(x))
The TRS R consists of the following rules:
b.0(b.0(b.0(b.0(B.1(x))))) → c.0(b.0(b.0(b.0(B.1(x)))))
b.0(b.0(B.0(x))) → c.0(A.0(x))
A.1(x0) → A.0(x0)
c.0(b.0(B.1(x))) → B.1(x)
a.1(x0) → a.0(x0)
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.0(A.0(x)))))
B.1(x0) → B.0(x0)
c.0(b.0(b.1(x))) → b.0(b.1(a.0(b.1(x))))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.1(a.0(B.0(x))))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.0(x)))))))
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(b.0(B.0(x))))
b.0(b.0(b.0(x))) → c.1(a.0(b.0(c.0(x))))
a.1(a.1(a.0(c.0(x)))) → c.1(a.0(x))
c.0(b.0(B.0(x))) → B.0(x)
b.0(b.0(c.1(a.0(x)))) → c.0(b.0(b.0(b.0(x))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(b.0(B.1(x))))
b.0(b.0(c.0(A.0(x)))) → C.0(x)
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(B.0(x)))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.1(x)))))))
c.0(b.0(B.0(x))) → b.0(B.0(x))
a.1(a.1(a.0(C.1(x)))) → C.1(x)
c.0(b.0(b.0(x))) → b.0(b.1(a.0(b.0(x))))
a.1(a.1(a.0(b.0(b.0(B.0(x)))))) → c.1(a.1(a.0(B.0(x))))
a.1(a.1(a.0(b.0(b.0(c.0(A.1(x))))))) → c.1(a.0(B.1(x)))
b.1(x0) → b.0(x0)
b.0(b.0(B.1(x))) → c.0(A.1(x))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.1(a.0(B.1(x))))))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.0(A.0(x)))))
a.1(a.1(a.0(C.0(x)))) → C.0(x)
b.0(b.0(b.0(b.0(B.0(x))))) → c.0(b.0(b.0(b.0(B.0(x)))))
b.0(b.0(c.1(a.1(x)))) → c.0(b.0(b.0(b.1(x))))
a.1(a.1(a.0(b.0(b.0(B.1(x)))))) → c.1(a.1(a.0(B.1(x))))
a.1(a.1(a.0(C.0(x)))) → c.0(A.0(x))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(B.1(x)))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.0(A.0(x))))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.1(a.0(B.1(x))))))
b.0(b.0(B.0(x))) → C.0(x)
c.0(b.0(B.1(x))) → b.0(B.1(x))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.0(A.1(x))))))
a.1(a.1(a.0(b.0(b.0(c.0(A.0(x))))))) → c.1(a.0(B.0(x)))
c.1(x0) → c.0(x0)
b.0(b.0(c.0(A.1(x)))) → C.1(x)
C.1(x0) → C.0(x0)
a.1(a.1(a.0(c.1(x)))) → c.1(a.1(x))
b.0(b.0(B.1(x))) → C.1(x)
a.1(a.1(a.0(C.1(x)))) → c.0(A.1(x))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.1(a.0(B.0(x))))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.0(A.1(x)))))
b.0(b.0(b.1(x))) → c.1(a.0(b.0(c.1(x))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.0(A.1(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 18 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ SemLabProof2
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1.0(b.0(b.0(x))) → C1.0(x)
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.0(x0))))))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(x)
B1.0(b.0(c.1(a.1(x)))) → B1.0(b.0(b.1(x)))
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.0(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(x0)))))))
B1.0(b.0(c.1(a.0(x)))) → C1.0(b.0(b.0(b.0(x))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.0(x0))))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.0(x0)))))))
B1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.0(b.1(a.0(b.1(x0)))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(b.0(x)))
B1.0(b.0(c.1(a.1(x)))) → C1.0(b.0(b.0(b.1(x))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.0(x0)))))))
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.1(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.1(x0))))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.1(x0))))))
C1.0(b.0(b.0(b.0(c.1(a.1(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.1(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.0(x0))))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(x))
B1.0(b.0(b.1(x))) → C1.0(x)
B1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.0(b.1(a.0(b.0(x0)))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.1(x0)))))))
B1.0(b.0(c.1(a.1(x)))) → B1.0(x)
B1.0(b.0(c.1(a.1(x)))) → B1.0(b.1(x))
The TRS R consists of the following rules:
b.0(b.0(b.0(b.0(B.1(x))))) → c.0(b.0(b.0(b.0(B.1(x)))))
b.0(b.0(B.0(x))) → c.0(A.0(x))
A.1(x0) → A.0(x0)
c.0(b.0(B.1(x))) → B.1(x)
a.1(x0) → a.0(x0)
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.0(A.0(x)))))
B.1(x0) → B.0(x0)
c.0(b.0(b.1(x))) → b.0(b.1(a.0(b.1(x))))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.1(a.0(B.0(x))))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.0(x)))))))
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(b.0(B.0(x))))
b.0(b.0(b.0(x))) → c.1(a.0(b.0(c.0(x))))
a.1(a.1(a.0(c.0(x)))) → c.1(a.0(x))
c.0(b.0(B.0(x))) → B.0(x)
b.0(b.0(c.1(a.0(x)))) → c.0(b.0(b.0(b.0(x))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(b.0(B.1(x))))
b.0(b.0(c.0(A.0(x)))) → C.0(x)
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(B.0(x)))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.1(x)))))))
c.0(b.0(B.0(x))) → b.0(B.0(x))
a.1(a.1(a.0(C.1(x)))) → C.1(x)
c.0(b.0(b.0(x))) → b.0(b.1(a.0(b.0(x))))
a.1(a.1(a.0(b.0(b.0(B.0(x)))))) → c.1(a.1(a.0(B.0(x))))
a.1(a.1(a.0(b.0(b.0(c.0(A.1(x))))))) → c.1(a.0(B.1(x)))
b.1(x0) → b.0(x0)
b.0(b.0(B.1(x))) → c.0(A.1(x))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.1(a.0(B.1(x))))))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.0(A.0(x)))))
a.1(a.1(a.0(C.0(x)))) → C.0(x)
b.0(b.0(b.0(b.0(B.0(x))))) → c.0(b.0(b.0(b.0(B.0(x)))))
b.0(b.0(c.1(a.1(x)))) → c.0(b.0(b.0(b.1(x))))
a.1(a.1(a.0(b.0(b.0(B.1(x)))))) → c.1(a.1(a.0(B.1(x))))
a.1(a.1(a.0(C.0(x)))) → c.0(A.0(x))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(B.1(x)))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.0(A.0(x))))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.1(a.0(B.1(x))))))
b.0(b.0(B.0(x))) → C.0(x)
c.0(b.0(B.1(x))) → b.0(B.1(x))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.0(A.1(x))))))
a.1(a.1(a.0(b.0(b.0(c.0(A.0(x))))))) → c.1(a.0(B.0(x)))
c.1(x0) → c.0(x0)
b.0(b.0(c.0(A.1(x)))) → C.1(x)
C.1(x0) → C.0(x0)
a.1(a.1(a.0(c.1(x)))) → c.1(a.1(x))
b.0(b.0(B.1(x))) → C.1(x)
a.1(a.1(a.0(C.1(x)))) → c.0(A.1(x))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.1(a.0(B.0(x))))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.0(A.1(x)))))
b.0(b.0(b.1(x))) → c.1(a.0(b.0(c.1(x))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.0(A.1(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
B1.0(b.0(c.1(a.1(x)))) → B1.0(b.0(b.1(x)))
B1.0(b.0(c.1(a.1(x)))) → C1.0(b.0(b.0(b.1(x))))
C1.0(b.0(b.0(b.0(c.1(a.1(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.1(x0)))))))
B1.0(b.0(c.1(a.1(x)))) → B1.0(x)
B1.0(b.0(c.1(a.1(x)))) → B1.0(b.1(x))
The following rules are removed from R:
b.0(b.0(c.1(a.1(x)))) → c.0(b.0(b.0(b.1(x))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = x1
POL(A.1(x1)) = x1
POL(B.0(x1)) = x1
POL(B.1(x1)) = x1
POL(B1.0(x1)) = x1
POL(C.0(x1)) = x1
POL(C.1(x1)) = x1
POL(C1.0(x1)) = x1
POL(a.0(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
POL(c.1(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ SemLabProof2
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1.0(b.0(b.0(x))) → C1.0(x)
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.0(x0))))))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(x)
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.1(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.0(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.1(x0))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(x0)))))))
B1.0(b.0(c.1(a.0(x)))) → C1.0(b.0(b.0(b.0(x))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.0(x0))))))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.0(x0))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.0(x0)))))))
B1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.0(b.1(a.0(b.1(x0)))))
B1.0(b.0(b.1(x))) → C1.0(x)
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(x))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(b.0(x)))
B1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.0(b.1(a.0(b.0(x0)))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.0(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.1(x0)))))))
The TRS R consists of the following rules:
b.0(b.0(b.0(b.0(B.1(x))))) → c.0(b.0(b.0(b.0(B.1(x)))))
b.0(b.0(B.1(x))) → c.0(A.1(x))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.0(A.0(x)))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.1(a.0(B.1(x))))))
b.0(b.0(b.0(b.0(B.0(x))))) → c.0(b.0(b.0(b.0(B.0(x)))))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.0(A.0(x)))))
c.0(b.0(b.1(x))) → b.0(b.1(a.0(b.1(x))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.0(A.0(x))))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.1(a.0(B.1(x))))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(B.1(x)))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.1(a.0(B.0(x))))))
c.0(b.0(B.1(x))) → b.0(B.1(x))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.0(A.1(x))))))
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(b.0(B.0(x))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.0(x)))))))
c.1(x0) → c.0(x0)
b.0(b.0(b.0(x))) → c.1(a.0(b.0(c.0(x))))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.1(a.0(B.0(x))))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.0(A.1(x)))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(b.0(B.1(x))))
b.0(b.0(c.1(a.0(x)))) → c.0(b.0(b.0(b.0(x))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.0(A.1(x)))))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.1(x)))))))
b.0(b.0(b.1(x))) → c.1(a.0(b.0(c.1(x))))
c.0(b.0(b.0(x))) → b.0(b.1(a.0(b.0(x))))
b.0(b.0(B.0(x))) → c.0(A.0(x))
b.0(b.0(c.0(A.0(x)))) → C.0(x)
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(B.0(x)))
b.0(b.0(B.0(x))) → C.0(x)
b.0(b.0(c.0(A.1(x)))) → C.1(x)
b.0(b.0(B.1(x))) → C.1(x)
C.1(x0) → C.0(x0)
c.0(b.0(B.0(x))) → B.0(x)
c.0(b.0(B.0(x))) → b.0(B.0(x))
b.1(x0) → b.0(x0)
B.1(x0) → B.0(x0)
c.0(b.0(B.1(x))) → B.1(x)
A.1(x0) → A.0(x0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
C.1(x0) → C.0(x0)
B.1(x0) → B.0(x0)
A.1(x0) → A.0(x0)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = x1
POL(A.1(x1)) = 1 + x1
POL(B.0(x1)) = x1
POL(B.1(x1)) = 1 + x1
POL(B1.0(x1)) = x1
POL(C.0(x1)) = x1
POL(C.1(x1)) = 1 + x1
POL(C1.0(x1)) = x1
POL(a.0(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
POL(c.1(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ SemLabProof2
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1.0(b.0(b.0(x))) → C1.0(x)
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.0(x0))))))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(x)
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.1(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.0(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.1(x0))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(x0)))))))
B1.0(b.0(c.1(a.0(x)))) → C1.0(b.0(b.0(b.0(x))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.0(x0))))))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.0(x0))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.0(x0)))))))
B1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.0(b.1(a.0(b.1(x0)))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(b.0(x)))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(x))
B1.0(b.0(b.1(x))) → C1.0(x)
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.0(x0)))))))
B1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.0(b.1(a.0(b.0(x0)))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.1(x0)))))))
The TRS R consists of the following rules:
b.0(b.0(b.0(b.0(B.1(x))))) → c.0(b.0(b.0(b.0(B.1(x)))))
b.0(b.0(B.1(x))) → c.0(A.1(x))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.0(A.0(x)))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.1(a.0(B.1(x))))))
b.0(b.0(b.0(b.0(B.0(x))))) → c.0(b.0(b.0(b.0(B.0(x)))))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.0(A.0(x)))))
c.0(b.0(b.1(x))) → b.0(b.1(a.0(b.1(x))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.0(A.0(x))))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.1(a.0(B.1(x))))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(B.1(x)))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.1(a.0(B.0(x))))))
c.0(b.0(B.1(x))) → b.0(B.1(x))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.0(A.1(x))))))
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(b.0(B.0(x))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.0(x)))))))
c.1(x0) → c.0(x0)
b.0(b.0(b.0(x))) → c.1(a.0(b.0(c.0(x))))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.1(a.0(B.0(x))))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.0(A.1(x)))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(b.0(B.1(x))))
b.0(b.0(c.1(a.0(x)))) → c.0(b.0(b.0(b.0(x))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.0(A.1(x)))))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.1(x)))))))
b.0(b.0(b.1(x))) → c.1(a.0(b.0(c.1(x))))
c.0(b.0(b.0(x))) → b.0(b.1(a.0(b.0(x))))
b.0(b.0(B.0(x))) → c.0(A.0(x))
b.0(b.0(c.0(A.0(x)))) → C.0(x)
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(B.0(x)))
b.0(b.0(B.0(x))) → C.0(x)
b.0(b.0(c.0(A.1(x)))) → C.1(x)
b.0(b.0(B.1(x))) → C.1(x)
c.0(b.0(B.0(x))) → B.0(x)
c.0(b.0(B.0(x))) → b.0(B.0(x))
b.1(x0) → b.0(x0)
c.0(b.0(B.1(x))) → B.1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
b.0(b.0(c.0(A.1(x)))) → C.1(x)
b.0(b.0(B.1(x))) → C.1(x)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = x1
POL(A.1(x1)) = 1 + x1
POL(B.0(x1)) = x1
POL(B.1(x1)) = 1 + x1
POL(B1.0(x1)) = x1
POL(C.0(x1)) = x1
POL(C.1(x1)) = x1
POL(C1.0(x1)) = x1
POL(a.0(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
POL(c.1(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ SemLabProof2
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1.0(b.0(b.0(x))) → C1.0(x)
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.0(x0))))))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(x)
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.1(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.0(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.1(x0))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(x0)))))))
B1.0(b.0(c.1(a.0(x)))) → C1.0(b.0(b.0(b.0(x))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.0(x0))))))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.0(x0))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.0(x0)))))))
B1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.0(b.1(a.0(b.1(x0)))))
B1.0(b.0(b.1(x))) → C1.0(x)
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(x))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(b.0(x)))
B1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.0(b.1(a.0(b.0(x0)))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.0(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.1(x0)))))))
The TRS R consists of the following rules:
b.0(b.0(b.0(b.0(B.1(x))))) → c.0(b.0(b.0(b.0(B.1(x)))))
b.0(b.0(B.1(x))) → c.0(A.1(x))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.0(A.0(x)))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.1(a.0(B.1(x))))))
b.0(b.0(b.0(b.0(B.0(x))))) → c.0(b.0(b.0(b.0(B.0(x)))))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.0(A.0(x)))))
c.0(b.0(b.1(x))) → b.0(b.1(a.0(b.1(x))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.0(A.0(x))))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.1(a.0(B.1(x))))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(B.1(x)))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.1(a.0(B.0(x))))))
c.0(b.0(B.1(x))) → b.0(B.1(x))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.0(A.1(x))))))
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(b.0(B.0(x))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.0(x)))))))
c.1(x0) → c.0(x0)
b.0(b.0(b.0(x))) → c.1(a.0(b.0(c.0(x))))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.1(a.0(B.0(x))))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.0(A.1(x)))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(b.0(B.1(x))))
b.0(b.0(c.1(a.0(x)))) → c.0(b.0(b.0(b.0(x))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.0(A.1(x)))))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.1(x)))))))
b.0(b.0(b.1(x))) → c.1(a.0(b.0(c.1(x))))
c.0(b.0(b.0(x))) → b.0(b.1(a.0(b.0(x))))
b.0(b.0(B.0(x))) → c.0(A.0(x))
b.0(b.0(c.0(A.0(x)))) → C.0(x)
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(B.0(x)))
b.0(b.0(B.0(x))) → C.0(x)
c.0(b.0(B.0(x))) → B.0(x)
c.0(b.0(B.0(x))) → b.0(B.0(x))
b.1(x0) → b.0(x0)
c.0(b.0(B.1(x))) → B.1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
b.0(b.0(c.0(A.0(x)))) → C.0(x)
b.0(b.0(B.0(x))) → C.0(x)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = 1 + x1
POL(A.1(x1)) = x1
POL(B.0(x1)) = 1 + x1
POL(B.1(x1)) = x1
POL(B1.0(x1)) = 1 + x1
POL(C.0(x1)) = x1
POL(C1.0(x1)) = 1 + x1
POL(a.0(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
POL(c.1(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ SemLabProof2
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1.0(b.0(b.0(x))) → C1.0(x)
C1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.0(x0))))))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(x)
C1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.1(x0)))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.1(a.0(c.1(a.0(b.0(c.0(x0)))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.1(x0))))))
B1.0(b.0(b.0(c.0(b.0(B.1(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.1(x0)))))))
C1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(B.1(x0))))))))
C1.0(b.0(b.0(b.0(c.1(a.0(x0)))))) → B1.0(b.1(a.0(c.0(b.0(b.0(b.0(x0)))))))
B1.0(b.0(c.1(a.0(x)))) → C1.0(b.0(b.0(b.0(x))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.1(a.0(B.0(x0))))))))
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.0(A.0(x0))))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.0(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.0(x0)))))))
B1.0(b.0(b.0(b.0(b.1(x0))))) → B1.0(b.0(b.1(a.0(b.1(x0)))))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(b.0(x)))
B1.0(b.0(c.1(a.0(x)))) → B1.0(b.0(x))
B1.0(b.0(b.1(x))) → C1.0(x)
B1.0(b.0(b.0(c.0(b.0(B.0(x0)))))) → B1.0(b.0(b.1(a.0(b.1(a.0(B.0(x0)))))))
B1.0(b.0(b.0(b.0(b.0(x0))))) → B1.0(b.0(b.1(a.0(b.0(x0)))))
B1.0(b.0(b.0(b.0(c.0(b.0(B.1(x0))))))) → B1.0(b.0(b.1(a.0(b.0(b.0(A.1(x0)))))))
The TRS R consists of the following rules:
b.0(b.0(b.0(b.0(B.1(x))))) → c.0(b.0(b.0(b.0(B.1(x)))))
b.0(b.0(B.1(x))) → c.0(A.1(x))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.0(A.0(x)))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.1(a.0(B.1(x))))))
b.0(b.0(b.0(b.0(B.0(x))))) → c.0(b.0(b.0(b.0(B.0(x)))))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.0(A.0(x)))))
c.0(b.0(b.1(x))) → b.0(b.1(a.0(b.1(x))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.0(A.0(x))))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.1(a.0(B.1(x))))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(B.1(x)))
b.0(c.0(b.0(B.0(x)))) → c.1(a.0(b.0(c.1(a.0(B.0(x))))))
c.0(b.0(B.1(x))) → b.0(B.1(x))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.0(A.1(x))))))
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(b.0(B.0(x))))
c.0(b.0(c.0(b.0(B.0(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.0(x)))))))
c.1(x0) → c.0(x0)
b.0(b.0(b.0(x))) → c.1(a.0(b.0(c.0(x))))
c.0(c.0(b.0(B.0(x)))) → b.0(b.1(a.0(b.1(a.0(B.0(x))))))
b.0(c.0(b.0(B.1(x)))) → c.1(a.0(b.0(c.0(A.1(x)))))
b.0(b.0(c.0(A.1(x)))) → c.0(b.0(b.0(B.1(x))))
b.0(b.0(c.1(a.0(x)))) → c.0(b.0(b.0(b.0(x))))
c.0(c.0(b.0(B.1(x)))) → b.0(b.1(a.0(b.0(A.1(x)))))
c.0(b.0(c.0(b.0(B.1(x))))) → b.0(b.1(a.0(b.0(b.1(a.0(B.1(x)))))))
b.0(b.0(b.1(x))) → c.1(a.0(b.0(c.1(x))))
c.0(b.0(b.0(x))) → b.0(b.1(a.0(b.0(x))))
b.0(b.0(B.0(x))) → c.0(A.0(x))
b.0(b.0(c.0(A.0(x)))) → c.0(b.0(B.0(x)))
c.0(b.0(B.0(x))) → B.0(x)
c.0(b.0(B.0(x))) → b.0(B.0(x))
b.1(x0) → b.0(x0)
c.0(b.0(B.1(x))) → B.1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used.
Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(A(x0))))))
B1(b(b(c(b(B(x0)))))) → B1(b(b(a(b(a(B(x0)))))))
B1(b(c(a(x)))) → C1(b(b(b(x))))
C1(b(b(b(b(b(B(x0))))))) → B1(b(a(c(b(b(b(B(x0))))))))
B1(b(b(b(b(x0))))) → B1(b(b(a(b(x0)))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(A(x0)))))))
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(b(x))) → C1(x)
C1(b(b(b(b(x0))))) → B1(b(a(c(a(b(c(x0)))))))
B1(b(b(b(c(b(B(x0))))))) → B1(b(b(a(b(b(a(B(x0))))))))
C1(b(b(b(c(a(x0)))))) → B1(b(a(c(b(b(b(x0)))))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 10 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(x)))) → B1(b(b(x)))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(c(a(x)))) → B1(b(b(x))) at position [0] we obtained the following new rules:
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(A(x0)))))))
B1(b(c(a(c(A(x0)))))) → B1(c(b(b(B(x0)))))
B1(b(c(a(c(A(x0)))))) → B1(c(b(B(x0))))
B1(b(c(a(B(x0))))) → B1(c(A(x0)))
B1(b(c(a(b(B(x0)))))) → B1(b(c(A(x0))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(a(B(x0))))))))
B1(b(c(a(b(x0))))) → B1(c(a(b(c(x0)))))
B1(b(c(a(b(b(B(x0))))))) → B1(c(b(b(b(B(x0))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(B(x0)))))
B1(b(c(a(b(b(x0)))))) → B1(b(c(a(b(c(x0))))))
B1(b(c(a(b(b(b(B(x0)))))))) → B1(b(c(b(b(b(B(x0)))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(b(B(x0))))))
B1(b(c(a(b(c(a(x0))))))) → B1(b(c(b(b(b(x0))))))
B1(b(c(a(c(a(x0)))))) → B1(c(b(b(b(x0)))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(A(x0)))))))
B1(b(c(a(c(A(x0)))))) → B1(c(b(b(B(x0)))))
B1(b(c(a(c(A(x0)))))) → B1(c(b(B(x0))))
B1(b(c(a(B(x0))))) → B1(c(A(x0)))
B1(b(c(a(b(B(x0)))))) → B1(b(c(A(x0))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(a(B(x0))))))))
B1(b(c(a(b(x0))))) → B1(c(a(b(c(x0)))))
B1(b(c(a(b(b(B(x0))))))) → B1(c(b(b(b(B(x0))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(B(x0)))))
B1(b(c(a(b(b(x0)))))) → B1(b(c(a(b(c(x0))))))
B1(b(c(a(b(b(b(B(x0)))))))) → B1(b(c(b(b(b(B(x0)))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(b(B(x0))))))
B1(b(c(a(b(c(a(x0))))))) → B1(b(c(b(b(b(x0))))))
B1(b(c(a(c(a(x0)))))) → B1(c(b(b(b(x0)))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(A(x0)))))))
B1(b(c(a(c(A(x0)))))) → B1(c(b(b(B(x0)))))
B1(b(c(a(c(A(x0)))))) → B1(c(b(B(x0))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(a(B(x0))))))))
B1(b(c(a(b(b(B(x0))))))) → B1(c(b(b(b(B(x0))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(B(x0)))))
B1(b(c(a(b(b(x0)))))) → B1(b(c(a(b(c(x0))))))
B1(b(c(a(b(b(b(B(x0)))))))) → B1(b(c(b(b(b(B(x0)))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(b(B(x0))))))
B1(b(c(a(b(c(a(x0))))))) → B1(b(c(b(b(b(x0))))))
B1(b(c(a(c(a(x0)))))) → B1(c(b(b(b(x0)))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(c(a(c(A(x0)))))) → B1(c(b(b(B(x0))))) at position [0] we obtained the following new rules:
B1(b(c(a(c(A(x0)))))) → B1(c(c(A(x0))))
B1(b(c(a(c(A(y0)))))) → B1(b(b(a(b(B(y0))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(A(x0)))))))
B1(b(c(a(c(A(x0)))))) → B1(c(b(B(x0))))
B1(b(c(a(c(A(x0)))))) → B1(c(c(A(x0))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(a(B(x0))))))))
B1(b(c(a(b(b(B(x0))))))) → B1(c(b(b(b(B(x0))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(B(x0)))))
B1(b(c(a(b(b(x0)))))) → B1(b(c(a(b(c(x0))))))
B1(b(c(a(b(b(b(B(x0)))))))) → B1(b(c(b(b(b(B(x0)))))))
B1(b(c(a(c(A(y0)))))) → B1(b(b(a(b(B(y0))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(b(B(x0))))))
B1(b(c(a(b(c(a(x0))))))) → B1(b(c(b(b(b(x0))))))
B1(b(c(a(c(a(x0)))))) → B1(c(b(b(b(x0)))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(B(x0)))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(A(x0)))))))
B1(b(c(a(b(b(x0)))))) → B1(b(c(a(b(c(x0))))))
B1(b(c(a(b(b(b(B(x0)))))))) → B1(b(c(b(b(b(B(x0)))))))
B1(b(c(a(c(A(x0)))))) → B1(c(b(B(x0))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(b(B(x0))))))
B1(b(c(a(b(c(a(x0))))))) → B1(b(c(b(b(b(x0))))))
B1(b(c(a(c(a(x0)))))) → B1(c(b(b(b(x0)))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(a(B(x0))))))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(b(b(B(x0))))))) → B1(c(b(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(c(a(c(A(x0)))))) → B1(c(b(B(x0)))) at position [0] we obtained the following new rules:
B1(b(c(a(c(A(x0)))))) → B1(B(x0))
B1(b(c(a(c(A(x0)))))) → B1(b(B(x0)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(A(x0)))))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(a(B(x0))))))))
B1(b(c(a(b(b(B(x0))))))) → B1(c(b(b(b(B(x0))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(B(x0)))))
B1(b(c(a(b(b(x0)))))) → B1(b(c(a(b(c(x0))))))
B1(b(c(a(c(A(x0)))))) → B1(b(B(x0)))
B1(b(c(a(c(A(x0)))))) → B1(B(x0))
B1(b(c(a(b(b(b(B(x0)))))))) → B1(b(c(b(b(b(B(x0)))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(b(B(x0))))))
B1(b(c(a(b(c(a(x0))))))) → B1(b(c(b(b(b(x0))))))
B1(b(c(a(c(a(x0)))))) → B1(c(b(b(b(x0)))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(B(x0)))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(A(x0)))))))
B1(b(c(a(b(b(x0)))))) → B1(b(c(a(b(c(x0))))))
B1(b(c(a(b(b(b(B(x0)))))))) → B1(b(c(b(b(b(B(x0)))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(b(B(x0))))))
B1(b(c(a(b(c(a(x0))))))) → B1(b(c(b(b(b(x0))))))
B1(b(c(a(c(a(x0)))))) → B1(c(b(b(b(x0)))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(a(B(x0))))))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(b(b(B(x0))))))) → B1(c(b(b(b(B(x0))))))
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(c(a(b(b(B(x0))))))) → B1(c(b(b(b(B(x0)))))) at position [0] we obtained the following new rules:
B1(b(c(a(b(b(B(x0))))))) → B1(c(b(c(A(x0)))))
B1(b(c(a(b(b(B(y0))))))) → B1(b(b(a(b(b(B(y0)))))))
B1(b(c(a(b(b(B(y0))))))) → B1(c(c(a(b(c(B(y0)))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(A(x0)))))))
B1(b(c(a(b(b(B(y0))))))) → B1(b(b(a(b(b(B(y0)))))))
B1(b(c(a(b(b(B(y0))))))) → B1(c(c(a(b(c(B(y0)))))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(a(B(x0))))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(B(x0)))))
B1(b(c(a(b(b(x0)))))) → B1(b(c(a(b(c(x0))))))
B1(b(c(a(b(b(b(B(x0)))))))) → B1(b(c(b(b(b(B(x0)))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(b(B(x0))))))
B1(b(c(a(b(c(a(x0))))))) → B1(b(c(b(b(b(x0))))))
B1(b(c(a(c(a(x0)))))) → B1(c(b(b(b(x0)))))
B1(b(c(a(b(b(B(x0))))))) → B1(c(b(c(A(x0)))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(B(x0)))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(A(x0)))))))
B1(b(c(a(b(b(x0)))))) → B1(b(c(a(b(c(x0))))))
B1(b(c(a(b(b(b(B(x0)))))))) → B1(b(c(b(b(b(B(x0)))))))
B1(b(c(a(b(c(A(x0))))))) → B1(b(c(b(b(B(x0))))))
B1(b(c(a(b(c(a(x0))))))) → B1(b(c(b(b(b(x0))))))
B1(b(c(a(c(a(x0)))))) → B1(c(b(b(b(x0)))))
B1(b(c(a(c(b(B(x0))))))) → B1(b(c(a(b(c(a(B(x0))))))))
B1(b(c(a(x)))) → B1(x)
B1(b(c(a(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(b(b(x))) → b(b(a(b(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(c(A(x)))) → c(b(B(x)))
c(b(B(x))) → b(B(x))
b(b(c(A(x)))) → c(b(b(B(x))))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(B(x))) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(c(x)))) → A1(x)
The TRS R consists of the following rules:
b(b(c(a(x)))) → c(b(b(b(x))))
c(b(b(x))) → b(b(a(b(x))))
a(a(a(c(x)))) → c(a(x))
b(b(b(x))) → c(a(b(c(x))))
b(b(c(A(x)))) → C(x)
b(b(c(A(x)))) → c(b(B(x)))
a(a(a(b(b(B(x)))))) → c(a(a(B(x))))
a(a(a(b(b(c(A(x))))))) → c(a(B(x)))
c(b(c(b(B(x))))) → b(b(a(b(b(a(B(x)))))))
c(b(c(b(B(x))))) → b(b(a(b(b(A(x))))))
b(b(B(x))) → C(x)
b(b(c(A(x)))) → c(b(b(B(x))))
c(c(b(B(x)))) → b(b(a(b(a(B(x))))))
b(b(b(b(B(x))))) → c(b(b(b(B(x)))))
b(b(B(x))) → c(A(x))
a(a(a(C(x)))) → c(A(x))
c(b(B(x))) → B(x)
b(c(b(B(x)))) → c(a(b(c(a(B(x))))))
c(b(B(x))) → b(B(x))
b(c(b(B(x)))) → c(a(b(c(A(x)))))
c(c(b(B(x)))) → b(b(a(b(A(x)))))
a(a(a(C(x)))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.