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.