Termination w.r.t. Q proof of TRS/secret06tpa06.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))
p₁(s₁(x)) -> x
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
f₃(0, y, z) -> max₂(y, z)
f₃(x, 0, z) -> max₂(x, z)
f₃(x, y, 0) -> max₂(x, 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))
p₁(s₁(x)) -> x
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
f₃(0, y, z) -> max₂(y, z)
f₃(x, 0, z) -> max₂(x, z)
f₃(x, y, 0) -> max₂(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:

F₃(s₁(x), s₁(y), s₁(z)) -> MIN₂(s₁(x), max₂(s₁(y), s₁(z)))
F₃(x, y, 0) -> MAX₂(x, y)
F₃(x, 0, z) -> MAX₂(x, z)
F₃(0, y, z) -> MAX₂(y, z)
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
F₃(s₁(x), s₁(y), s₁(z)) -> MAX₂(s₁(y), s₁(z))
MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
F₃(s₁(x), s₁(y), s₁(z)) -> MIN₂(s₁(y), s₁(z))
F₃(s₁(x), s₁(y), s₁(z)) -> P₁(min₂(s₁(x), max₂(s₁(y), s₁(z))))
F₃(s₁(x), s₁(y), s₁(z)) -> MAX₂(s₁(x), max₂(s₁(y), s₁(z)))
F₃(s₁(x), s₁(y), s₁(z)) -> MIN₂(s₁(x), min₂(s₁(y), s₁(z)))

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))
p₁(s₁(x)) -> x
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
f₃(0, y, z) -> max₂(y, z)
f₃(x, 0, z) -> max₂(x, z)
f₃(x, y, 0) -> max₂(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:

F₃(s₁(x), s₁(y), s₁(z)) -> MIN₂(s₁(x), max₂(s₁(y), s₁(z)))
F₃(x, y, 0) -> MAX₂(x, y)
F₃(x, 0, z) -> MAX₂(x, z)
F₃(0, y, z) -> MAX₂(y, z)
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
F₃(s₁(x), s₁(y), s₁(z)) -> MAX₂(s₁(y), s₁(z))
MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
F₃(s₁(x), s₁(y), s₁(z)) -> MIN₂(s₁(y), s₁(z))
F₃(s₁(x), s₁(y), s₁(z)) -> P₁(min₂(s₁(x), max₂(s₁(y), s₁(z))))
F₃(s₁(x), s₁(y), s₁(z)) -> MAX₂(s₁(x), max₂(s₁(y), s₁(z)))
F₃(s₁(x), s₁(y), s₁(z)) -> MIN₂(s₁(x), min₂(s₁(y), s₁(z)))

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))
p₁(s₁(x)) -> x
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
f₃(0, y, z) -> max₂(y, z)
f₃(x, 0, z) -> max₂(x, z)
f₃(x, y, 0) -> max₂(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 3 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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))
p₁(s₁(x)) -> x
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
f₃(0, y, z) -> max₂(y, z)
f₃(x, 0, z) -> max₂(x, z)
f₃(x, y, 0) -> max₂(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.

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)) = 3·x₁ + 3·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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))
p₁(s₁(x)) -> x
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
f₃(0, y, z) -> max₂(y, z)
f₃(x, 0, z) -> max₂(x, z)
f₃(x, y, 0) -> max₂(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

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))
p₁(s₁(x)) -> x
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
f₃(0, y, z) -> max₂(y, z)
f₃(x, 0, z) -> max₂(x, z)
f₃(x, y, 0) -> max₂(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)) = 3·x₁ + 3·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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))
p₁(s₁(x)) -> x
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
f₃(0, y, z) -> max₂(y, z)
f₃(x, 0, z) -> max₂(x, z)
f₃(x, y, 0) -> max₂(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

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

F₃(s₁(x), s₁(y), s₁(z)) -> F₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))

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))
p₁(s₁(x)) -> x
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(max₂(s₁(x), max₂(s₁(y), s₁(z))), p₁(min₂(s₁(x), max₂(s₁(y), s₁(z)))), min₂(s₁(x), min₂(s₁(y), s₁(z))))
f₃(0, y, z) -> max₂(y, z)
f₃(x, 0, z) -> max₂(x, z)
f₃(x, y, 0) -> max₂(x, y)

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