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(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(x1)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(x1)
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:
B(c(x1)) → A(x1)
A(b(b(x1))) → A(x1)
A(b(b(x1))) → B(b(b(c(a(x1)))))
A(b(b(x1))) → B(b(c(a(x1))))
A(b(b(x1))) → B(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(x1)) → A(x1)
A(b(b(x1))) → A(x1)
A(b(b(x1))) → B(b(b(c(a(x1)))))
A(b(b(x1))) → B(b(c(a(x1))))
A(b(b(x1))) → B(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(x1))) → B(b(b(c(a(x1))))) at position [0] we obtained the following new rules:
A(b(b(x0))) → B(b(b(c(x0))))
A(b(b(b(b(x0))))) → B(b(b(c(b(b(b(c(a(x0)))))))))
A(b(b(y0))) → B(b(a(a(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x0))) → B(b(b(c(x0))))
A(b(b(x1))) → A(x1)
B(c(x1)) → A(x1)
A(b(b(x1))) → B(b(c(a(x1))))
A(b(b(x1))) → B(c(a(x1)))
A(b(b(y0))) → B(b(a(a(y0))))
A(b(b(b(b(x0))))) → B(b(b(c(b(b(b(c(a(x0)))))))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(x1))) → B(b(c(a(x1)))) at position [0] we obtained the following new rules:
A(b(b(b(b(x0))))) → B(b(c(b(b(b(c(a(x0))))))))
A(b(b(y0))) → B(a(a(y0)))
A(b(b(x0))) → B(b(c(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(b(b(x0))))) → B(b(c(b(b(b(c(a(x0))))))))
B(c(x1)) → A(x1)
A(b(b(x1))) → A(x1)
A(b(b(x0))) → B(b(b(c(x0))))
A(b(b(y0))) → B(a(a(y0)))
A(b(b(x0))) → B(b(c(x0)))
A(b(b(x1))) → B(c(a(x1)))
A(b(b(b(b(x0))))) → B(b(b(c(b(b(b(c(a(x0)))))))))
A(b(b(y0))) → B(b(a(a(y0))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(x1)
A(b(b(b(b(x0))))) → B(b(c(b(b(b(c(a(x0))))))))
B(c(x1)) → A(x1)
A(b(b(x1))) → A(x1)
A(b(b(x0))) → B(b(b(c(x0))))
A(b(b(y0))) → B(a(a(y0)))
A(b(b(x0))) → B(b(c(x0)))
A(b(b(x1))) → B(c(a(x1)))
A(b(b(b(b(x0))))) → B(b(b(c(b(b(b(c(a(x0)))))))))
A(b(b(y0))) → B(b(a(a(y0))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(x1)
A(b(b(b(b(x0))))) → B(b(c(b(b(b(c(a(x0))))))))
B(c(x1)) → A(x1)
A(b(b(x1))) → A(x1)
A(b(b(x0))) → B(b(b(c(x0))))
A(b(b(y0))) → B(a(a(y0)))
A(b(b(x0))) → B(b(c(x0)))
A(b(b(x1))) → B(c(a(x1)))
A(b(b(b(b(x0))))) → B(b(b(c(b(b(b(c(a(x0)))))))))
A(b(b(y0))) → B(b(a(a(y0))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(b(x))) → b(b(b(c(a(x)))))
b(c(x)) → a(x)
A(b(b(b(b(x))))) → B(b(c(b(b(b(c(a(x))))))))
B(c(x)) → A(x)
A(b(b(x))) → A(x)
A(b(b(x))) → B(b(b(c(x))))
A(b(b(x))) → B(a(a(x)))
A(b(b(x))) → B(b(c(x)))
A(b(b(x))) → B(c(a(x)))
A(b(b(b(b(x))))) → B(b(b(c(b(b(b(c(a(x)))))))))
A(b(b(x))) → B(b(a(a(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(b(b(x))) → b(b(b(c(a(x)))))
b(c(x)) → a(x)
A(b(b(b(b(x))))) → B(b(c(b(b(b(c(a(x))))))))
B(c(x)) → A(x)
A(b(b(x))) → A(x)
A(b(b(x))) → B(b(b(c(x))))
A(b(b(x))) → B(a(a(x)))
A(b(b(x))) → B(b(c(x)))
A(b(b(x))) → B(c(a(x)))
A(b(b(b(b(x))))) → B(b(b(c(b(b(b(c(a(x)))))))))
A(b(b(x))) → B(b(a(a(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(b(x))) → b(b(b(c(a(x)))))
b(c(x)) → a(x)
A(b(b(b(b(x))))) → B(b(c(b(b(b(c(a(x))))))))
B(c(x)) → A(x)
A(b(b(x))) → A(x)
A(b(b(x))) → B(b(b(c(x))))
A(b(b(x))) → B(a(a(x)))
A(b(b(x))) → B(b(c(x)))
A(b(b(x))) → B(c(a(x)))
A(b(b(b(b(x))))) → B(b(b(c(b(b(b(c(a(x)))))))))
A(b(b(x))) → B(b(a(a(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(b(b(x))) → b(b(b(c(a(x)))))
b(c(x)) → a(x)
A(b(b(b(b(x))))) → B(b(c(b(b(b(c(a(x))))))))
B(c(x)) → A(x)
A(b(b(x))) → A(x)
A(b(b(x))) → B(b(b(c(x))))
A(b(b(x))) → B(a(a(x)))
A(b(b(x))) → B(b(c(x)))
A(b(b(x))) → B(c(a(x)))
A(b(b(b(b(x))))) → B(b(b(c(b(b(b(c(a(x)))))))))
A(b(b(x))) → B(b(a(a(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(b(A(x))) → A1(a(b(B(x))))
B1(b(a(x))) → C(b(b(b(x))))
B1(b(b(b(A(x))))) → B1(B(x))
B1(b(b(b(A(x))))) → B1(b(c(b(B(x)))))
B1(b(b(b(A(x))))) → B1(c(b(b(B(x)))))
B1(b(b(b(A(x))))) → C(b(b(b(c(b(b(B(x))))))))
B1(b(A(x))) → A1(a(B(x)))
B1(b(A(x))) → C(b(B(x)))
B1(b(b(b(A(x))))) → B1(b(b(c(b(b(B(x)))))))
B1(b(b(b(A(x))))) → C(b(b(b(c(b(B(x)))))))
B1(b(a(x))) → A1(c(b(b(b(x)))))
B1(b(A(x))) → C(B(x))
B1(b(A(x))) → C(b(b(B(x))))
B1(b(b(b(A(x))))) → A1(c(b(b(b(c(b(b(B(x)))))))))
B1(b(A(x))) → B1(b(B(x)))
B1(b(A(x))) → B1(B(x))
B1(b(b(b(A(x))))) → C(b(B(x)))
B1(b(A(x))) → A1(b(B(x)))
B1(b(A(x))) → A1(B(x))
C(b(x)) → A1(x)
B1(b(A(x))) → A1(c(B(x)))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(x))))) → B1(b(B(x)))
B1(b(a(x))) → B1(b(x))
B1(b(b(b(A(x))))) → B1(c(b(B(x))))
B1(b(b(b(A(x))))) → A1(c(b(b(b(c(b(B(x))))))))
B1(b(b(b(A(x))))) → B1(b(b(c(b(B(x))))))
B1(b(b(b(A(x))))) → B1(b(c(b(b(B(x))))))
B1(b(b(b(A(x))))) → C(b(b(B(x))))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(A(x))) → A1(a(b(B(x))))
B1(b(a(x))) → C(b(b(b(x))))
B1(b(b(b(A(x))))) → B1(B(x))
B1(b(b(b(A(x))))) → B1(b(c(b(B(x)))))
B1(b(b(b(A(x))))) → B1(c(b(b(B(x)))))
B1(b(b(b(A(x))))) → C(b(b(b(c(b(b(B(x))))))))
B1(b(A(x))) → A1(a(B(x)))
B1(b(A(x))) → C(b(B(x)))
B1(b(b(b(A(x))))) → B1(b(b(c(b(b(B(x)))))))
B1(b(b(b(A(x))))) → C(b(b(b(c(b(B(x)))))))
B1(b(a(x))) → A1(c(b(b(b(x)))))
B1(b(A(x))) → C(B(x))
B1(b(A(x))) → C(b(b(B(x))))
B1(b(b(b(A(x))))) → A1(c(b(b(b(c(b(b(B(x)))))))))
B1(b(A(x))) → B1(b(B(x)))
B1(b(A(x))) → B1(B(x))
B1(b(b(b(A(x))))) → C(b(B(x)))
B1(b(A(x))) → A1(b(B(x)))
B1(b(A(x))) → A1(B(x))
C(b(x)) → A1(x)
B1(b(A(x))) → A1(c(B(x)))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(x))))) → B1(b(B(x)))
B1(b(a(x))) → B1(b(x))
B1(b(b(b(A(x))))) → B1(c(b(B(x))))
B1(b(b(b(A(x))))) → A1(c(b(b(b(c(b(B(x))))))))
B1(b(b(b(A(x))))) → B1(b(b(c(b(B(x))))))
B1(b(b(b(A(x))))) → B1(b(c(b(b(B(x))))))
B1(b(b(b(A(x))))) → C(b(b(B(x))))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(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 21 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(x))))) → B1(b(b(c(b(b(B(x)))))))
B1(b(b(b(A(x))))) → B1(c(b(B(x))))
B1(b(b(b(A(x))))) → B1(b(c(b(B(x)))))
B1(b(b(b(A(x))))) → B1(c(b(b(B(x)))))
B1(b(b(b(A(x))))) → B1(b(b(c(b(B(x))))))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(x))))) → B1(b(c(b(b(B(x))))))
B1(b(a(x))) → B1(b(x))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(A(x))))) → B1(b(b(c(b(B(x)))))) at position [0] we obtained the following new rules:
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(x))))) → B1(c(b(B(x))))
B1(b(b(b(A(x))))) → B1(b(b(c(b(b(B(x)))))))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(b(b(A(x))))) → B1(b(c(b(B(x)))))
B1(b(b(b(A(x))))) → B1(c(b(b(B(x)))))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(x))))) → B1(b(c(b(b(B(x))))))
B1(b(a(x))) → B1(b(x))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(A(x))))) → B1(b(c(b(B(x))))) at position [0] we obtained the following new rules:
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(x))))) → B1(b(b(c(b(b(B(x)))))))
B1(b(b(b(A(x))))) → B1(c(b(B(x))))
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(b(b(A(x))))) → B1(c(b(b(B(x)))))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(x))))) → B1(b(c(b(b(B(x))))))
B1(b(a(x))) → B1(b(x))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(A(x))))) → B1(c(b(B(x)))) at position [0] we obtained the following new rules:
B1(b(b(b(A(y0))))) → B1(a(B(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(x))))) → B1(b(b(c(b(b(B(x)))))))
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(b(b(A(x))))) → B1(c(b(b(B(x)))))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(x))))) → B1(b(c(b(b(B(x))))))
B1(b(a(x))) → B1(b(x))
B1(b(b(b(A(y0))))) → B1(a(B(y0)))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(A(x))))) → B1(b(b(c(b(b(B(x))))))) at position [0] we obtained the following new rules:
B1(b(b(b(A(y0))))) → B1(b(b(a(b(B(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(b(b(A(x))))) → B1(c(b(b(B(x)))))
B1(b(b(b(A(y0))))) → B1(b(b(a(b(B(y0))))))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(x))))) → B1(b(c(b(b(B(x))))))
B1(b(a(x))) → B1(b(x))
B1(b(b(b(A(y0))))) → B1(a(B(y0)))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(A(x))))) → B1(b(c(b(b(B(x)))))) at position [0] we obtained the following new rules:
B1(b(b(b(A(y0))))) → B1(b(a(b(B(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(a(b(B(y0)))))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(b(b(A(x))))) → B1(c(b(b(B(x)))))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(y0))))) → B1(b(b(a(b(B(y0))))))
B1(b(a(x))) → B1(b(x))
B1(b(b(b(A(y0))))) → B1(a(B(y0)))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(A(x))))) → B1(c(b(b(B(x))))) at position [0] we obtained the following new rules:
B1(b(b(b(A(y0))))) → B1(a(b(B(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(a(b(B(y0)))))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(b(b(A(y0))))) → B1(a(b(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(b(a(b(B(y0))))))
B1(b(a(x))) → B1(x)
B1(b(a(x))) → B1(b(x))
B1(b(b(b(A(y0))))) → B1(a(B(y0)))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(A(y0))))) → B1(a(B(y0))) at position [0] we obtained the following new rules:
B1(b(b(b(A(y0))))) → B1(B(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(a(b(B(y0)))))
B1(b(b(b(A(y0))))) → B1(B(y0))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(b(b(A(y0))))) → B1(a(b(B(y0))))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(y0))))) → B1(b(b(a(b(B(y0))))))
B1(b(a(x))) → B1(b(x))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(a(b(B(y0)))))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(b(b(A(y0))))) → B1(a(b(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(b(a(b(B(y0))))))
B1(b(a(x))) → B1(x)
B1(b(a(x))) → B1(b(x))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(A(y0))))) → B1(a(b(B(y0)))) at position [0] we obtained the following new rules:
B1(b(b(b(A(y0))))) → B1(b(B(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(a(b(B(y0)))))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(a(x))) → B1(x)
B1(b(b(b(A(y0))))) → B1(b(b(a(b(B(y0))))))
B1(b(b(b(A(y0))))) → B1(b(B(y0)))
B1(b(a(x))) → B1(b(x))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(A(y0))))) → B1(b(a(B(y0))))
B1(b(b(b(A(y0))))) → B1(b(a(b(B(y0)))))
B1(b(b(b(A(y0))))) → B1(b(b(a(B(y0)))))
B1(b(b(b(A(y0))))) → B1(b(b(a(b(B(y0))))))
B1(b(a(x))) → B1(x)
B1(b(a(x))) → B1(b(x))
B1(b(a(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(B(x))))))))
c(B(x)) → A(x)
b(b(A(x))) → A(x)
b(b(A(x))) → c(b(b(B(x))))
b(b(A(x))) → a(a(B(x)))
b(b(A(x))) → c(b(B(x)))
b(b(A(x))) → a(c(B(x)))
b(b(b(b(A(x))))) → a(c(b(b(b(c(b(b(B(x)))))))))
b(b(A(x))) → a(a(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(b(x1))) → b(b(b(c(a(x1)))))
b(c(x1)) → a(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(b(b(b(x)))))
c(b(x)) → a(x)
Q is empty.