Termination w.r.t. Q proof of TRS/secret06tpa10.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:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

Q is empty.

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

*¹₂(x, s₁(y)) -> +¹₂(x, *₂(x, y))
F₂(s₁(x), s₁(y)) -> MIN₂(s₁(x), s₁(y))
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
*¹₂(x, s₁(y)) -> *¹₂(x, y)
F₂(s₁(x), s₁(y)) -> -¹₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y)))
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₂(s₁(x), s₁(y)) -> *¹₂(s₁(x), s₁(y))
+¹₂(s₁(x), y) -> +¹₂(x, y)
MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)
F₂(s₁(x), s₁(y)) -> MAX₂(s₁(x), s₁(y))
F₂(s₁(x), s₁(y)) -> F₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

*¹₂(x, s₁(y)) -> +¹₂(x, *₂(x, y))
F₂(s₁(x), s₁(y)) -> MIN₂(s₁(x), s₁(y))
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
*¹₂(x, s₁(y)) -> *¹₂(x, y)
F₂(s₁(x), s₁(y)) -> -¹₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y)))
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₂(s₁(x), s₁(y)) -> *¹₂(s₁(x), s₁(y))
+¹₂(s₁(x), y) -> +¹₂(x, y)
MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)
F₂(s₁(x), s₁(y)) -> MAX₂(s₁(x), s₁(y))
F₂(s₁(x), s₁(y)) -> F₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

Q is empty.
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.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-¹₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

+¹₂(s₁(x), y) -> +¹₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

Q is empty.
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.

+¹₂(s₁(x), y) -> +¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+¹₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

*¹₂(x, s₁(y)) -> *¹₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

Q is empty.
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.

*¹₂(x, s₁(y)) -> *¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*¹₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

Q is empty.
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.

MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MAX₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

Q is empty.
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.

MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MIN₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₂(s₁(x), s₁(y)) -> F₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
p₁(s₁(x)) -> x
f₂(s₁(x), s₁(y)) -> f₂(-₂(min₂(s₁(x), s₁(y)), max₂(s₁(x), s₁(y))), *₂(s₁(x), s₁(y)))

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