Termination w.r.t. Q proof of TRS/Cimefact-hard.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → 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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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

IFFACT(x, true) → *¹(x, fact(-(x, s(0))))
*¹(x, s(y)) → +¹(*(x, y), x)
+¹(x, s(y)) → +¹(x, y)
*¹(x, s(y)) → *¹(x, y)
FACT(x) → IFFACT(x, ge(x, s(s(0))))
-¹(s(x), s(y)) → -¹(x, y)
IFFACT(x, true) → FACT(-(x, s(0)))
IFFACT(x, true) → -¹(x, s(0))
FACT(x) → GE(x, s(s(0)))
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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:

IFFACT(x, true) → *¹(x, fact(-(x, s(0))))
*¹(x, s(y)) → +¹(*(x, y), x)
+¹(x, s(y)) → +¹(x, y)
*¹(x, s(y)) → *¹(x, y)
FACT(x) → IFFACT(x, ge(x, s(s(0))))
-¹(s(x), s(y)) → -¹(x, y)
IFFACT(x, true) → FACT(-(x, s(0)))
IFFACT(x, true) → -¹(x, s(0))
FACT(x) → GE(x, s(s(0)))
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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)
IFFACT(x, true) → *¹(x, fact(-(x, s(0))))
+¹(x, s(y)) → +¹(x, y)
*¹(x, s(y)) → *¹(x, y)
-¹(s(x), s(y)) → -¹(x, y)
FACT(x) → IFFACT(x, ge(x, s(s(0))))
IFFACT(x, true) → FACT(-(x, s(0)))
IFFACT(x, true) → -¹(x, s(0))
FACT(x) → GE(x, s(s(0)))
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

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

                  ↳ QDP

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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

                  ↳ QDP

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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.

GE(s(x), s(y)) → GE(x, y)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

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

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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

                  ↳ 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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

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

                  ↳ QDP

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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

                  ↳ 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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

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

                  ↳ QDP

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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

                  ↳ QDP

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

FACT(x) → IFFACT(x, ge(x, s(s(0))))
IFFACT(x, true) → FACT(-(x, s(0)))

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
fact(x) → iffact(x, ge(x, s(s(0))))
iffact(x, true) → *(x, fact(-(x, s(0))))
iffact(x, false) → s(0)

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))
*(x0, 0)
*(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
fact(x0)
iffact(x0, true)
iffact(x0, false)

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