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
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
The remaining pairs can at least be oriented weakly.

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(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))
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(APP2(x1, x2)) = 2·x1   
POL(app2(x1, x2)) = 2 + 2·x1·x2   
POL(cons) = 0   
POL(hd) = 0   
POL(map) = 2   
POL(nil) = 0   

The following usable rules [14] were oriented:

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))
app2(app2(f, 0), n) -> app2(app2(hd, app2(app2(map, f), app2(app2(cons, 0), nil))), n)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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(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)

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.