Termination w.r.t. Q proof of TRS/secret05tpa4.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
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x

The TRS R 2 is

f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The signature Sigma is {f}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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, *(x, y))
*¹(x, s(y)) → *¹(x, y)
-¹(s(x), s(y)) → -¹(x, y)
+¹(s(x), y) → +¹(x, y)
F(s(x)) → *¹(s(x), s(x))
F(s(x)) → P(*(s(x), s(x)))
F(s(x)) → -¹(p(*(s(x), s(x))), *(s(x), s(x)))
F(s(x)) → F(-(p(*(s(x), s(x))), *(s(x), s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

*¹(x, s(y)) → +¹(x, *(x, y))
*¹(x, s(y)) → *¹(x, y)
-¹(s(x), s(y)) → -¹(x, y)
+¹(s(x), y) → +¹(x, y)
F(s(x)) → *¹(s(x), s(x))
F(s(x)) → P(*(s(x), s(x)))
F(s(x)) → -¹(p(*(s(x), s(x))), *(s(x), s(x)))
F(s(x)) → F(-(p(*(s(x), s(x))), *(s(x), s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

*¹(x, s(y)) → +¹(x, *(x, y))
*¹(x, s(y)) → *¹(x, y)
-¹(s(x), s(y)) → -¹(x, y)
+¹(s(x), y) → +¹(x, y)
F(s(x)) → *¹(s(x), s(x))
F(s(x)) → -¹(p(*(s(x), s(x))), *(s(x), s(x)))
F(s(x)) → P(*(s(x), s(x)))
F(s(x)) → F(-(p(*(s(x), s(x))), *(s(x), s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ 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
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

F(s(x)) → F(-(p(*(s(x), s(x))), *(s(x), s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x)) → f(-(p(*(s(x), s(x))), *(s(x), s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0))

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