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:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(X))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(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:
F(g(f(X))) → F(g(X))
F(f(X)) → F(g(f(g(f(X)))))
F(f(X)) → F(g(f(X)))
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(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:
F(g(f(X))) → F(g(X))
F(f(X)) → F(g(f(g(f(X)))))
F(f(X)) → F(g(f(X)))
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(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
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(f(X)) → F(g(f(g(f(X)))))
F(g(f(X))) → F(g(X))
F(f(X)) → F(g(f(X)))
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(g(f(X))) → F(g(X))
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(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.
F(g(f(X))) → F(g(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1) = x1
g(x1) = x1
f(x1) = f(x1)
Recursive Path Order [2].
Precedence:
trivial
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(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.