Termination w.r.t. Q proof of TRS/SK902.41.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:

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

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:

NORM₁(g₂(x, y)) -> NORM₁(x)
F₂(x, g₂(y, z)) -> F₂(x, y)
REM₂(g₂(x, y), s₁(z)) -> REM₂(x, z)

The TRS R consists of the following rules:

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

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:

NORM₁(g₂(x, y)) -> NORM₁(x)
F₂(x, g₂(y, z)) -> F₂(x, y)
REM₂(g₂(x, y), s₁(z)) -> REM₂(x, z)

The TRS R consists of the following rules:

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

REM₂(g₂(x, y), s₁(z)) -> REM₂(x, z)

The TRS R consists of the following rules:

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

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.

REM₂(g₂(x, y), s₁(z)) -> REM₂(x, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(REM₂(x₁, x₂)) = 3·x₁ + x₁·x₂ + 2·x₂
POL(g₂(x₁, x₂)) = 2 + 2·x₁
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

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

          ↳ QDP

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

F₂(x, g₂(y, z)) -> F₂(x, y)

The TRS R consists of the following rules:

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

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.

F₂(x, g₂(y, z)) -> F₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = x₂
POL(g₂(x₁, x₂)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

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

          ↳ QDP

            ↳ QDPOrderProof

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

NORM₁(g₂(x, y)) -> NORM₁(x)

The TRS R consists of the following rules:

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

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.

NORM₁(g₂(x, y)) -> NORM₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(NORM₁(x₁)) = 3·x₁ + 3·x₁²
POL(g₂(x₁, x₂)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

norm₁(nil) -> 0
norm₁(g₂(x, y)) -> s₁(norm₁(x))
f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
rem₂(nil, y) -> nil
rem₂(g₂(x, y), 0) -> g₂(x, y)
rem₂(g₂(x, y), s₁(z)) -> rem₂(x, z)

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.