Termination w.r.t. Q proof of TRS/secret06tpa04.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:

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

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

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

Q is empty.

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

min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
min(0, y) → 0

The TRS R 2 is

f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(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:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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)) → P(twice(min(x, y)))
F(s(x), s(y)) → TWICE(min(x, y))
F(s(x), s(y)) → MIN(s(x), s(y))
MIN(s(x), s(y)) → MIN(x, y)
F(s(x), s(y)) → MIN(x, y)
F(s(x), s(y)) → -¹(max(s(x), s(y)), min(s(x), s(y)))
-¹(s(x), s(y)) → -¹(x, y)
F(s(x), s(y)) → F(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))
MAX(s(x), s(y)) → MAX(x, y)
F(s(x), s(y)) → MAX(s(x), s(y))
TWICE(s(x)) → TWICE(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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)) → P(twice(min(x, y)))
F(s(x), s(y)) → TWICE(min(x, y))
F(s(x), s(y)) → MIN(s(x), s(y))
MIN(s(x), s(y)) → MIN(x, y)
F(s(x), s(y)) → MIN(x, y)
F(s(x), s(y)) → -¹(max(s(x), s(y)), min(s(x), s(y)))
-¹(s(x), s(y)) → -¹(x, y)
F(s(x), s(y)) → F(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))
MAX(s(x), s(y)) → MAX(x, y)
F(s(x), s(y)) → MAX(s(x), s(y))
TWICE(s(x)) → TWICE(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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)) → P(twice(min(x, y)))
F(s(x), s(y)) → MIN(s(x), s(y))
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)) → -¹(max(s(x), s(y)), min(s(x), s(y)))
F(s(x), s(y)) → MIN(x, y)
F(s(x), s(y)) → F(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))
MAX(s(x), s(y)) → MAX(x, y)
F(s(x), s(y)) → MAX(s(x), s(y))
TWICE(s(x)) → TWICE(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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 5 SCCs with 6 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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)

Lexicographic Path Order [19].
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

                  ↳ QDP

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

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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

                  ↳ 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:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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)

Lexicographic Path Order [19].
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

                  ↳ QDP

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

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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

                  ↳ QDP

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

MAX(s(x), s(y)) → MAX(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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.

MAX(s(x), s(y)) → MAX(x, y)

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

Lexicographic Path Order [19].
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

                  ↳ QDP

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

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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

                    ↳ QDPOrderProof

                  ↳ 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:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(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

                  ↳ QDP

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

F(s(x), s(y)) → F(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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