Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(b(a(x))) → a(b(b(x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(b(a(x))) → a(b(b(x)))
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A(a(x)) → B(x)
B(b(a(x))) → A(b(b(x)))
B(b(a(x))) → B(x)
B(b(a(x))) → B(b(x))
A(a(x)) → B(b(x))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(b(a(x))) → a(b(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
A(a(x)) → B(x)
B(b(a(x))) → A(b(b(x)))
B(b(a(x))) → B(x)
B(b(a(x))) → B(b(x))
A(a(x)) → B(b(x))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(b(a(x))) → a(b(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
A(a(x)) → B(x)
B(b(a(x))) → A(b(b(x)))
B(b(a(x))) → B(x)
B(b(a(x))) → B(b(x))
A(a(x)) → B(b(x))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(b(a(x))) → a(b(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
A(a(x)) → B(x)
B(b(a(x))) → A(b(b(x)))
B(b(a(x))) → B(x)
B(b(a(x))) → B(b(x))
A(a(x)) → B(b(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A(x1) = x1
a(x1) = a(x1)
B(x1) = B(x1)
b(x1) = x1
Recursive Path Order [2].
Precedence:
a1 > B1
The following usable rules [14] were oriented:
a(a(x)) → b(b(x))
b(b(a(x))) → a(b(b(x)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(b(a(x))) → a(b(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.