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:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

The set Q consists of the following terms:

ap(ap(ff, x0), x0)


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

AP(ap(ff, x), x) → AP(x, ap(ff, x))
AP(ap(ff, x), x) → AP(cons, x)
AP(ap(ff, x), x) → AP(ap(x, ap(ff, x)), ap(ap(cons, x), nil))
AP(ap(ff, x), x) → AP(ap(cons, x), nil)

The TRS R consists of the following rules:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

The set Q consists of the following terms:

ap(ap(ff, x0), x0)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

AP(ap(ff, x), x) → AP(x, ap(ff, x))
AP(ap(ff, x), x) → AP(cons, x)
AP(ap(ff, x), x) → AP(ap(x, ap(ff, x)), ap(ap(cons, x), nil))
AP(ap(ff, x), x) → AP(ap(cons, x), nil)

The TRS R consists of the following rules:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

The set Q consists of the following terms:

ap(ap(ff, x0), x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 4 less nodes.