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:
ap2(ap2(ff, x), x) -> ap2(ap2(x, ap2(ff, x)), ap2(ap2(cons, x), nil))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
ap2(ap2(ff, x), x) -> ap2(ap2(x, ap2(ff, x)), ap2(ap2(cons, x), nil))
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(ff, x), x) -> ap2(ap2(x, ap2(ff, x)), ap2(ap2(cons, x), nil))
The set Q consists of the following terms:
ap2(ap2(ff, x0), x0)
Q DP problem:
The TRS P consists of the following rules:
AP2(ap2(ff, x), x) -> AP2(ap2(cons, x), nil)
AP2(ap2(ff, x), x) -> AP2(ap2(x, ap2(ff, x)), ap2(ap2(cons, x), nil))
AP2(ap2(ff, x), x) -> AP2(x, ap2(ff, x))
AP2(ap2(ff, x), x) -> AP2(cons, x)
The TRS R consists of the following rules:
ap2(ap2(ff, x), x) -> ap2(ap2(x, ap2(ff, x)), ap2(ap2(cons, x), nil))
The set Q consists of the following terms:
ap2(ap2(ff, x0), x0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
AP2(ap2(ff, x), x) -> AP2(ap2(cons, x), nil)
AP2(ap2(ff, x), x) -> AP2(ap2(x, ap2(ff, x)), ap2(ap2(cons, x), nil))
AP2(ap2(ff, x), x) -> AP2(x, ap2(ff, x))
AP2(ap2(ff, x), x) -> AP2(cons, x)
The TRS R consists of the following rules:
ap2(ap2(ff, x), x) -> ap2(ap2(x, ap2(ff, x)), ap2(ap2(cons, x), nil))
The set Q consists of the following terms:
ap2(ap2(ff, x0), x0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 4 less nodes.