Termination w.r.t. Q proof of /tmp/tmpmqmit9/016.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

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(neq, app(s, x)), app(s, y)) → APP(neq, x)
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(cons, y), app(app(filter, f), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(filter, f)
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(filtersub, app(f, y)), f)
APP(app(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)
NONZERO → APP(neq, 0)
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
NONZERO → APP(filter, app(neq, 0))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(filter, f)
APP(app(filter, f), app(app(cons, y), ys)) → APP(filtersub, app(f, y))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

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(neq, app(s, x)), app(s, y)) → APP(neq, x)
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(cons, y), app(app(filter, f), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(filter, f)
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(filtersub, app(f, y)), f)
APP(app(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)
NONZERO → APP(neq, 0)
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
NONZERO → APP(filter, app(neq, 0))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(filter, f)
APP(app(filter, f), app(app(cons, y), ys)) → APP(filtersub, app(f, y))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

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 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

APP(app(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(app(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (1/4)x_1
POL(app(x₁, x₂)) = 1 + (2)x_2
POL(s) = 0
POL(neq) = 0

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)

The remaining pairs can at least be oriented weakly.

APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (1/2)x_1
POL(cons) = 0
POL(true) = 0
POL(false) = 0
POL(app(x₁, x₂)) = 1/4 + x_2
POL(s) = 0
POL(filtersub) = 0
POL(neq) = 0
POL(0) = 0
POL(filter) = 0
POL(nil) = 0

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The remaining pairs can at least be oriented weakly.

APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))

Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (1/2)x_2
POL(cons) = 0
POL(true) = 0
POL(false) = 0
POL(app(x₁, x₂)) = 4 + x_2
POL(s) = 0
POL(filtersub) = 0
POL(neq) = 0
POL(0) = 0
POL(filter) = 0
POL(nil) = 0

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.