Termination w.r.t. Q proof of TRS/SK902.18.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:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
+(x0, 0)
+(x0, s(x1))

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

+¹(x, s(y)) → +¹(x, y)
SUM(s(x)) → SUM(x)
SUM(s(x)) → +¹(sum(x), s(x))

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
+(x0, 0)
+(x0, s(x1))

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:

+¹(x, s(y)) → +¹(x, y)
SUM(s(x)) → SUM(x)
SUM(s(x)) → +¹(sum(x), s(x))

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
+(x0, 0)
+(x0, s(x1))

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:

+¹(x, s(y)) → +¹(x, y)
SUM(s(x)) → SUM(x)
SUM(s(x)) → +¹(sum(x), s(x))

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
+(x0, 0)
+(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
+(x0, 0)
+(x0, s(x1))

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.

+¹(x, s(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = +¹(x2)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

[+^1₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
+(x0, 0)
+(x0, s(x1))

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

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

SUM(s(x)) → SUM(x)

The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
+(x0, 0)
+(x0, s(x1))

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.

SUM(s(x)) → SUM(x)

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

sum(0)
sum(s(x0))
+(x0, 0)
+(x0, s(x1))

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