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(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → 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(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → 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:
A(b(x1)) → B(b(a(c(x1))))
B(b(x1)) → A(x1)
A(b(x1)) → A(c(x1))
A(b(x1)) → C(x1)
A(b(x1)) → B(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(a(c(x1))))
B(b(x1)) → A(x1)
A(b(x1)) → A(c(x1))
A(b(x1)) → C(x1)
A(b(x1)) → B(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → 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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(a(c(x1))))
B(b(x1)) → A(x1)
A(b(x1)) → A(c(x1))
A(b(x1)) → B(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → A(c(x1)) at position [0] we obtained the following new rules:
A(b(c(x0))) → A(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(a(c(x1))))
B(b(x1)) → A(x1)
A(b(x1)) → B(a(c(x1)))
A(b(c(x0))) → A(x0)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → B(a(c(x1))) at position [0] we obtained the following new rules:
A(b(c(x0))) → B(a(x0))
A(b(y0)) → B(c(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(a(c(x1))))
A(b(y0)) → B(c(y0))
A(b(c(x0))) → B(a(x0))
B(b(x1)) → A(x1)
A(b(c(x0))) → A(x0)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(y0)) → B(c(y0)) at position [0] we obtained the following new rules:
A(b(c(x0))) → B(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(a(c(x1))))
A(b(c(x0))) → B(a(x0))
A(b(c(x0))) → B(x0)
B(b(x1)) → A(x1)
A(b(c(x0))) → A(x0)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → 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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → x1
A(b(x1)) → B(b(a(c(x1))))
A(b(c(x0))) → B(a(x0))
A(b(c(x0))) → B(x0)
B(b(x1)) → A(x1)
A(b(c(x0))) → A(x0)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → x1
A(b(x1)) → B(b(a(c(x1))))
A(b(c(x0))) → B(a(x0))
A(b(c(x0))) → B(x0)
B(b(x1)) → A(x1)
A(b(c(x0))) → A(x0)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → x
b(A(x)) → c(a(b(B(x))))
c(b(A(x))) → a(B(x))
c(b(A(x))) → B(x)
b(B(x)) → A(x)
c(b(A(x))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → x
b(A(x)) → c(a(b(B(x))))
c(b(A(x))) → a(B(x))
c(b(A(x))) → B(x)
b(B(x)) → A(x)
c(b(A(x))) → 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(a(x)) → B1(b(x))
B1(a(x)) → B1(x)
B1(A(x)) → B1(B(x))
B1(A(x)) → C(a(b(B(x))))
B1(b(x)) → A1(x)
B1(A(x)) → A1(b(B(x)))
C(b(A(x))) → A1(B(x))
B1(a(x)) → A1(b(b(x)))
B1(a(x)) → C(a(b(b(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → x
b(A(x)) → c(a(b(B(x))))
c(b(A(x))) → a(B(x))
c(b(A(x))) → B(x)
b(B(x)) → A(x)
c(b(A(x))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(b(x))
B1(a(x)) → B1(x)
B1(A(x)) → B1(B(x))
B1(A(x)) → C(a(b(B(x))))
B1(b(x)) → A1(x)
B1(A(x)) → A1(b(B(x)))
C(b(A(x))) → A1(B(x))
B1(a(x)) → A1(b(b(x)))
B1(a(x)) → C(a(b(b(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → x
b(A(x)) → c(a(b(B(x))))
c(b(A(x))) → a(B(x))
c(b(A(x))) → B(x)
b(B(x)) → A(x)
c(b(A(x))) → A(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 7 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(b(x))
B1(a(x)) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → x
b(A(x)) → c(a(b(B(x))))
c(b(A(x))) → a(B(x))
c(b(A(x))) → B(x)
b(B(x)) → A(x)
c(b(A(x))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(x)) → B1(b(x)) at position [0] we obtained the following new rules:
B1(a(B(x0))) → B1(A(x0))
B1(a(a(x0))) → B1(c(a(b(b(x0)))))
B1(a(A(x0))) → B1(c(a(b(B(x0)))))
B1(a(b(x0))) → B1(a(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(B(x0))) → B1(A(x0))
B1(a(x)) → B1(x)
B1(a(b(x0))) → B1(a(x0))
B1(a(a(x0))) → B1(c(a(b(b(x0)))))
B1(a(A(x0))) → B1(c(a(b(B(x0)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → x
b(A(x)) → c(a(b(B(x))))
c(b(A(x))) → a(B(x))
c(b(A(x))) → B(x)
b(B(x)) → A(x)
c(b(A(x))) → A(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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(a(b(x0))) → B1(a(x0))
B1(a(a(x0))) → B1(c(a(b(b(x0)))))
B1(a(A(x0))) → B1(c(a(b(B(x0)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → x
b(A(x)) → c(a(b(B(x))))
c(b(A(x))) → a(B(x))
c(b(A(x))) → B(x)
b(B(x)) → A(x)
c(b(A(x))) → A(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(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → 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(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(b(x1)) → a(x1)
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → 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(a(x)) → c(a(b(b(x))))
b(b(x)) → a(x)
c(c(x)) → x
Q is empty.