Termination w.r.t. Q proof of TRS/SK902.23.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:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

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:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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), x)
FLOOP(s(x), y) → *¹(s(x), y)
FAC(s(x)) → FAC(x)
+¹(x, s(y)) → +¹(x, y)
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
FAC(s(x)) → *¹(s(x), fac(x))
*¹(x, s(y)) → *¹(x, y)
FAC(0) → 1¹

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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), x)
FLOOP(s(x), y) → *¹(s(x), y)
FAC(s(x)) → FAC(x)
+¹(x, s(y)) → +¹(x, y)
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
FAC(s(x)) → *¹(s(x), fac(x))
*¹(x, s(y)) → *¹(x, y)
FAC(0) → 1¹

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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), x)
FAC(s(x)) → FAC(x)
FLOOP(s(x), y) → *¹(s(x), y)
FAC(s(x)) → *¹(s(x), fac(x))
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
+¹(x, s(y)) → +¹(x, y)
*¹(x, s(y)) → *¹(x, y)
FAC(0) → 1¹

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ 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:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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(x1, x2)
s(x1) = s(x1)

Lexicographic path order with status [19].
Precedence:

s₁ > +^1₂

Status:

s₁: multiset
+^1₂: [1,2]

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

                  ↳ QDP

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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

                  ↳ 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:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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(x1, x2)
s(x1) = s(x1)

Lexicographic path order with status [19].
Precedence:

s₁ > *^1₂

Status:

*^1₂: [1,2]
s₁: multiset

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

                  ↳ QDP

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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

                  ↳ QDP

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

FLOOP(s(x), y) → FLOOP(x, *(s(x), y))

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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.

FLOOP(s(x), y) → FLOOP(x, *(s(x), y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FLOOP(x1, x2) = FLOOP(x1)
s(x1) = s(x1)
*(x1, x2) = *
0 = 0
+(x1, x2) = +

Lexicographic path order with status [19].
Precedence:

FLOOP₁ > s₁
* > s₁
0 > s₁
+ > s₁

Status:

FLOOP₁: [1]
+: []
s₁: [1]
0: multiset
*: []

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

                  ↳ QDP

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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

                  ↳ QDP

                    ↳ QDPOrderProof

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

FAC(s(x)) → FAC(x)

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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.

FAC(s(x)) → FAC(x)

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

Lexicographic path order with status [19].
Precedence:

s₁ > FAC₁

Status:

FAC₁: [1]
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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