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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, y))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, 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:

F₂(s₁(x), s₁(y)) -> TWICE₁(min₂(x, y))
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
F₂(s₁(x), s₁(y)) -> MIN₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₂(s₁(x), s₁(y)) -> -¹₂(x, min₂(x, y))
F₂(s₁(x), s₁(y)) -> F₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, y))))
F₂(s₁(x), s₁(y)) -> F₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
TWICE₁(s₁(x)) -> TWICE₁(x)
F₂(s₁(x), s₁(y)) -> -¹₂(y, min₂(x, y))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, 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:

F₂(s₁(x), s₁(y)) -> TWICE₁(min₂(x, y))
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
F₂(s₁(x), s₁(y)) -> MIN₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₂(s₁(x), s₁(y)) -> -¹₂(x, min₂(x, y))
F₂(s₁(x), s₁(y)) -> F₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, y))))
F₂(s₁(x), s₁(y)) -> F₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
TWICE₁(s₁(x)) -> TWICE₁(x)
F₂(s₁(x), s₁(y)) -> -¹₂(y, min₂(x, y))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, y))))

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TWICE₁(s₁(x)) -> TWICE₁(x)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, 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.

TWICE₁(s₁(x)) -> TWICE₁(x)

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

POL(TWICE₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, 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

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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, 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)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, 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

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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, 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)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, 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

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

F₂(s₁(x), s₁(y)) -> F₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
F₂(s₁(x), s₁(y)) -> F₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, y))))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
twice₁(0) -> 0
twice₁(s₁(x)) -> s₁(s₁(twice₁(x)))
f₂(s₁(x), s₁(y)) -> f₂(-₂(y, min₂(x, y)), s₁(twice₁(min₂(x, y))))
f₂(s₁(x), s₁(y)) -> f₂(-₂(x, min₂(x, y)), s₁(twice₁(min₂(x, y))))

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