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(b(x)) → b(b(a(x)))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
Q is empty.
The TRS is overlay and locally confluent. By [15] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
The set Q consists of the following terms:
a(b(x0))
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(b(x)) → A(x)
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
The set Q consists of the following terms:
a(b(x0))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
The set Q consists of the following terms:
a(b(x0))
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(b(x)) → A(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A(x1) = x1
b(x1) = b(x1)
Recursive Path Order [2].
Precedence:
trivial
The following usable rules [14] were oriented:
none
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
The set Q consists of the following terms:
a(b(x0))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.