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.