Termination w.r.t. Q proof of /tmp/tmptLzdNR/30.trs

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:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, app(app(:, x), y)), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), z))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, 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:

APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, app(app(:, x), y)), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), z))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 14 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), y)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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