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(f, 0), n) -> app2(app2(hd, app2(app2(map, f), app2(app2(cons, 0), nil))), n)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(f, 0), n) -> app2(app2(hd, app2(app2(map, f), app2(app2(cons, 0), nil))), n)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
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(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(f, 0), n) -> APP2(map, f)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(f, 0), n) -> APP2(app2(cons, 0), nil)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(f, 0), n) -> APP2(app2(map, f), app2(app2(cons, 0), nil))
APP2(app2(f, 0), n) -> APP2(app2(hd, app2(app2(map, f), app2(app2(cons, 0), nil))), n)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))
APP2(app2(f, 0), n) -> APP2(cons, 0)
APP2(app2(f, 0), n) -> APP2(hd, app2(app2(map, f), app2(app2(cons, 0), nil)))
The TRS R consists of the following rules:
app2(app2(f, 0), n) -> app2(app2(hd, app2(app2(map, f), app2(app2(cons, 0), nil))), n)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(f, 0), n) -> APP2(map, f)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(f, 0), n) -> APP2(app2(cons, 0), nil)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(f, 0), n) -> APP2(app2(map, f), app2(app2(cons, 0), nil))
APP2(app2(f, 0), n) -> APP2(app2(hd, app2(app2(map, f), app2(app2(cons, 0), nil))), n)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))
APP2(app2(f, 0), n) -> APP2(cons, 0)
APP2(app2(f, 0), n) -> APP2(hd, app2(app2(map, f), app2(app2(cons, 0), nil)))
The TRS R consists of the following rules:
app2(app2(f, 0), n) -> app2(app2(hd, app2(app2(map, f), app2(app2(cons, 0), nil))), n)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(f, 0), n) -> APP2(app2(cons, 0), nil)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(f, 0), n) -> APP2(app2(hd, app2(app2(map, f), app2(app2(cons, 0), nil))), n)
APP2(app2(f, 0), n) -> APP2(app2(map, f), app2(app2(cons, 0), nil))
The TRS R consists of the following rules:
app2(app2(f, 0), n) -> app2(app2(hd, app2(app2(map, f), app2(app2(cons, 0), nil))), n)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.