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