Termination w.r.t. Q proof of TRS/secret05tpa5.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)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

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)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

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)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))

The TRS R 2 is

f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

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)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(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:

F(s(x), s(y)) → TWICE(min(x, y))
MIN(s(x), s(y)) → MIN(x, y)
F(s(x), s(y)) → MIN(x, y)
-¹(s(x), s(y)) → -¹(x, y)
F(s(x), s(y)) → -¹(x, min(x, y))
F(s(x), s(y)) → F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) → F(-(y, min(x, y)), s(twice(min(x, y))))
TWICE(s(x)) → TWICE(x)
F(s(x), s(y)) → -¹(y, min(x, y))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(x0), s(x1))

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:

F(s(x), s(y)) → TWICE(min(x, y))
MIN(s(x), s(y)) → MIN(x, y)
F(s(x), s(y)) → MIN(x, y)
-¹(s(x), s(y)) → -¹(x, y)
F(s(x), s(y)) → -¹(x, min(x, y))
F(s(x), s(y)) → F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) → F(-(y, min(x, y)), s(twice(min(x, y))))
TWICE(s(x)) → TWICE(x)
F(s(x), s(y)) → -¹(y, min(x, y))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(x0), s(x1))

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:

F(s(x), s(y)) → TWICE(min(x, y))
MIN(s(x), s(y)) → MIN(x, y)
-¹(s(x), s(y)) → -¹(x, y)
F(s(x), s(y)) → MIN(x, y)
F(s(x), s(y)) → -¹(x, min(x, y))
F(s(x), s(y)) → F(-(y, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) → F(-(x, min(x, y)), s(twice(min(x, y))))
TWICE(s(x)) → TWICE(x)
F(s(x), s(y)) → -¹(y, min(x, y))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(x0), s(x1))

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:

TWICE(s(x)) → TWICE(x)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(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.

TWICE(s(x)) → TWICE(x)

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

Recursive path order with status [2].
Precedence:

trivial

Status:

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)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(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

  ↳ 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:

MIN(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(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.

MIN(s(x), s(y)) → MIN(x, y)

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

Recursive path order with status [2].
Precedence:

trivial

Status:

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)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(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

  ↳ 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)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(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(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 with status [2].
Precedence:

trivial

Status:

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)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(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

  ↳ 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), s(y)) → F(-(y, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) → F(-(x, min(x, y)), s(twice(min(x, y))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
f(s(x), s(y)) → f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) → f(-(x, min(x, y)), s(twice(min(x, y))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
twice(0)
twice(s(x0))
f(s(x0), s(x1))

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