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:
app2(app2(app2(rec, f), x), 0) -> x
app2(app2(app2(rec, f), x), app2(s, y)) -> app2(app2(f, app2(s, y)), app2(app2(app2(rec, f), x), y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(app2(rec, f), x), 0) -> x
app2(app2(app2(rec, f), x), app2(s, y)) -> app2(app2(f, app2(s, y)), app2(app2(app2(rec, f), x), y))
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(app2(rec, f), x), app2(s, y)) -> APP2(f, app2(s, y))
APP2(app2(app2(rec, f), x), app2(s, y)) -> APP2(app2(f, app2(s, y)), app2(app2(app2(rec, f), x), y))
APP2(app2(app2(rec, f), x), app2(s, y)) -> APP2(app2(app2(rec, f), x), y)
The TRS R consists of the following rules:
app2(app2(app2(rec, f), x), 0) -> x
app2(app2(app2(rec, f), x), app2(s, y)) -> app2(app2(f, app2(s, y)), app2(app2(app2(rec, f), x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(app2(rec, f), x), app2(s, y)) -> APP2(f, app2(s, y))
APP2(app2(app2(rec, f), x), app2(s, y)) -> APP2(app2(f, app2(s, y)), app2(app2(app2(rec, f), x), y))
APP2(app2(app2(rec, f), x), app2(s, y)) -> APP2(app2(app2(rec, f), x), y)
The TRS R consists of the following rules:
app2(app2(app2(rec, f), x), 0) -> x
app2(app2(app2(rec, f), x), app2(s, y)) -> app2(app2(f, app2(s, y)), app2(app2(app2(rec, f), x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.