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:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

NORM(g(x, y)) → NORM(x)
F(x, g(y, z)) → F(x, y)
REM(g(x, y), s(z)) → REM(x, z)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

NORM(g(x, y)) → NORM(x)
F(x, g(y, z)) → F(x, y)
REM(g(x, y), s(z)) → REM(x, z)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

NORM(g(x, y)) → NORM(x)
F(x, g(y, z)) → F(x, y)
REM(g(x, y), s(z)) → REM(x, z)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

REM(g(x, y), s(z)) → REM(x, z)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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.

REM(g(x, y), s(z)) → REM(x, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
REM(x1, x2) = REM(x2)
g(x1, x2) = g(x2)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

g₁ > REM₁
s₁ > REM₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

F(x, g(y, z)) → F(x, y)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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.

F(x, g(y, z)) → F(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2) = F(x2)
g(x1, x2) = g(x1, x2)

Recursive Path Order [2].
Precedence:

g₂ > F₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

NORM(g(x, y)) → NORM(x)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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.

NORM(g(x, y)) → NORM(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
NORM(x1) = NORM(x1)
g(x1, x2) = g(x1, x2)

Recursive Path Order [2].
Precedence:

g₂ > NORM₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

The set Q consists of the following terms:

norm(nil)
norm(g(x0, x1))
f(x0, nil)
f(x0, g(x1, x2))
rem(nil, x0)
rem(g(x0, x1), 0)
rem(g(x0, x1), s(x2))

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