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(g(X)) → g(f(f(X)))
f(h(X)) → h(g(X))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(X)) → g(f(f(X)))
f(h(X)) → h(g(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(X)) → F(X)
F(g(X)) → F(f(X))
The TRS R consists of the following rules:
f(g(X)) → g(f(f(X)))
f(h(X)) → h(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(g(X)) → F(X)
F(g(X)) → F(f(X))
The TRS R consists of the following rules:
f(g(X)) → g(f(f(X)))
f(h(X)) → h(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
F(g(X)) → F(X)
F(g(X)) → F(f(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(f(x1)) = x_1
POL(g(x1)) = 1/2 + (4)x_1
POL(h(x1)) = 0
POL(F(x1)) = x_1
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:
f(g(X)) → g(f(f(X)))
f(h(X)) → h(g(X))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(g(X)) → g(f(f(X)))
f(h(X)) → h(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.