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