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(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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))
app2(p, app2(s, x)) -> x
app2(app2(div, 0), app2(s, y)) -> 0
app2(app2(div, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(div, app2(app2(minus, x), y)), app2(s, y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))
app2(p, app2(s, x)) -> x
app2(app2(div, 0), app2(s, y)) -> 0
app2(app2(div, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(div, app2(app2(minus, x), y)), app2(s, 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(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(p, app2(s, y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(p, app2(s, x))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(minus, app2(p, app2(s, x)))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(div, app2(app2(minus, x), y))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(app2(div, app2(app2(minus, x), y)), app2(s, y))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(s, app2(app2(div, app2(app2(minus, x), y)), app2(s, y)))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))
app2(p, app2(s, x)) -> x
app2(app2(div, 0), app2(s, y)) -> 0
app2(app2(div, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(div, app2(app2(minus, x), y)), app2(s, y)))

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(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(p, app2(s, y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(p, app2(s, x))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(minus, app2(p, app2(s, x)))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(div, app2(app2(minus, x), y))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(app2(div, app2(app2(minus, x), y)), app2(s, y))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(s, app2(app2(div, app2(app2(minus, x), y)), app2(s, y)))
APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))
app2(p, app2(s, x)) -> x
app2(app2(div, 0), app2(s, y)) -> 0
app2(app2(div, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(div, app2(app2(minus, x), y)), app2(s, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 9 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))

The TRS R consists of the following rules:

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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))
app2(p, app2(s, x)) -> x
app2(app2(div, 0), app2(s, y)) -> 0
app2(app2(div, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(div, app2(app2(minus, x), y)), app2(s, y)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP2(app2(div, app2(s, x)), app2(s, y)) -> APP2(app2(div, app2(app2(minus, x), y)), app2(s, y))

The TRS R consists of the following rules:

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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))
app2(p, app2(s, x)) -> x
app2(app2(div, 0), app2(s, y)) -> 0
app2(app2(div, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(div, app2(app2(minus, x), y)), app2(s, y)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)

The TRS R consists of the following rules:

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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, app2(p, app2(s, x))), app2(p, app2(s, y)))
app2(p, app2(s, x)) -> x
app2(app2(div, 0), app2(s, y)) -> 0
app2(app2(div, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(div, app2(app2(minus, x), y)), app2(s, y)))

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