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:
ap2(ap2(g, x), y) -> y
ap2(f, x) -> ap2(f, app2(g, x))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
ap2(ap2(g, x), y) -> y
ap2(f, x) -> ap2(f, app2(g, x))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
ap2(ap2(g, x), y) -> y
ap2(f, x) -> ap2(f, app2(g, x))
The set Q consists of the following terms:
ap2(ap2(g, x0), x1)
ap2(f, x0)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
AP2(f, x) -> AP2(f, app2(g, x))
The TRS R consists of the following rules:
ap2(ap2(g, x), y) -> y
ap2(f, x) -> ap2(f, app2(g, x))
The set Q consists of the following terms:
ap2(ap2(g, x0), x1)
ap2(f, x0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
AP2(f, x) -> AP2(f, app2(g, x))
The TRS R consists of the following rules:
ap2(ap2(g, x), y) -> y
ap2(f, x) -> ap2(f, app2(g, x))
The set Q consists of the following terms:
ap2(ap2(g, x0), x1)
ap2(f, x0)
We have to consider all minimal (P,Q,R)-chains.