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:
h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(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:
F(g(f(x))) → F(f(x))
H(f(f(x))) → H(f(g(f(x))))
H(f(f(x))) → F(g(f(x)))
The TRS R consists of the following rules:
h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(g(f(x))) → F(f(x))
H(f(f(x))) → H(f(g(f(x))))
H(f(f(x))) → F(g(f(x)))
The TRS R consists of the following rules:
h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
H(f(f(x))) → H(f(g(f(x))))
The TRS R consists of the following rules:
h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.