Termination w.r.t. Q proof of TRS/secret05tpa2.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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(x, s₁(y)) -> -¹₂(s₁(y), x)
F₂(x, s₁(y)) -> -¹₂(x, s₁(y))
F₂(s₁(x), y) -> P₁(-₂(y, s₁(x)))
F₂(s₁(x), y) -> P₁(-₂(s₁(x), y))
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₂(s₁(x), y) -> F₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
F₂(x, s₁(y)) -> F₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))
F₂(x, s₁(y)) -> P₁(-₂(s₁(y), x))
F₂(x, s₁(y)) -> P₁(-₂(x, s₁(y)))
F₂(s₁(x), y) -> -¹₂(y, s₁(x))
F₂(s₁(x), y) -> -¹₂(s₁(x), y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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(x, s₁(y)) -> -¹₂(s₁(y), x)
F₂(x, s₁(y)) -> -¹₂(x, s₁(y))
F₂(s₁(x), y) -> P₁(-₂(y, s₁(x)))
F₂(s₁(x), y) -> P₁(-₂(s₁(x), y))
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₂(s₁(x), y) -> F₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
F₂(x, s₁(y)) -> F₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))
F₂(x, s₁(y)) -> P₁(-₂(s₁(y), x))
F₂(x, s₁(y)) -> P₁(-₂(x, s₁(y)))
F₂(s₁(x), y) -> -¹₂(y, s₁(x))
F₂(s₁(x), y) -> -¹₂(s₁(x), y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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 Order [17,21] with Interpretation:

POL( -¹₂(x₁, x₂) ) = max{0, x₂ - 2}

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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

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

F₂(s₁(x), y) -> F₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
F₂(x, s₁(y)) -> F₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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.

F₂(s₁(x), y) -> F₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))

The remaining pairs can at least be oriented weakly.

F₂(x, s₁(y)) -> F₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F₂(x₁, x₂) ) = x₁

POL( s₁(x₁) ) = x₁ + 1

POL( p₁(x₁) ) = max{0, x₁ - 1}

POL( -₂(x₁, x₂) ) = x₁

The following usable rules [14] were oriented:

-₂(x, 0) -> x
p₁(s₁(x)) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

F₂(x, s₁(y)) -> F₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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.

F₂(x, s₁(y)) -> F₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F₂(x₁, x₂) ) = max{0, x₂ - 2}

POL( s₁(x₁) ) = x₁ + 3

POL( p₁(x₁) ) = max{0, x₁ - 3}

POL( -₂(x₁, x₂) ) = x₁

The following usable rules [14] were oriented:

-₂(x, 0) -> x
p₁(s₁(x)) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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.