Termination w.r.t. Q proof of TRS/Koprowskigcd

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:

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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:

MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
GCD₃(x, s₁(y), s₁(z)) -> GCD₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
GCD₃(s₁(x), s₁(y), z) -> -¹₂(max₂(x, y), min₂(x, y))
GCD₃(s₁(x), y, s₁(z)) -> GCD₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
GCD₃(x, s₁(y), s₁(z)) -> -¹₂(max₂(y, z), min₂(y, z))
GCD₃(s₁(x), s₁(y), z) -> GCD₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
GCD₃(x, s₁(y), s₁(z)) -> MIN₂(y, z)
GCD₃(s₁(x), y, s₁(z)) -> MAX₂(x, z)
GCD₃(x, s₁(y), s₁(z)) -> MAX₂(y, z)
GCD₃(s₁(x), s₁(y), z) -> MIN₂(x, y)
GCD₃(s₁(x), s₁(y), z) -> MAX₂(x, y)
MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)
GCD₃(s₁(x), y, s₁(z)) -> MIN₂(x, z)
GCD₃(s₁(x), y, s₁(z)) -> -¹₂(max₂(x, z), min₂(x, z))

The TRS R consists of the following rules:

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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:

MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
GCD₃(x, s₁(y), s₁(z)) -> GCD₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
GCD₃(s₁(x), s₁(y), z) -> -¹₂(max₂(x, y), min₂(x, y))
GCD₃(s₁(x), y, s₁(z)) -> GCD₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
GCD₃(x, s₁(y), s₁(z)) -> -¹₂(max₂(y, z), min₂(y, z))
GCD₃(s₁(x), s₁(y), z) -> GCD₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
GCD₃(x, s₁(y), s₁(z)) -> MIN₂(y, z)
GCD₃(s₁(x), y, s₁(z)) -> MAX₂(x, z)
GCD₃(x, s₁(y), s₁(z)) -> MAX₂(y, z)
GCD₃(s₁(x), s₁(y), z) -> MIN₂(x, y)
GCD₃(s₁(x), s₁(y), z) -> MAX₂(x, y)
MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)
GCD₃(s₁(x), y, s₁(z)) -> MIN₂(x, z)
GCD₃(s₁(x), y, s₁(z)) -> -¹₂(max₂(x, z), min₂(x, z))

The TRS R consists of the following rules:

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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·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

          ↳ QDP

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

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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:

MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)

The TRS R consists of the following rules:

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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)) = 2·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

          ↳ QDP

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

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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:

MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)

The TRS R consists of the following rules:

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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·x₂
POL(s₁(x₁)) = 2 + 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:

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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

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

GCD₃(x, s₁(y), s₁(z)) -> GCD₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
GCD₃(s₁(x), y, s₁(z)) -> GCD₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
GCD₃(s₁(x), s₁(y), z) -> GCD₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)

The TRS R consists of the following rules:

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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.

GCD₃(x, s₁(y), s₁(z)) -> GCD₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
GCD₃(s₁(x), y, s₁(z)) -> GCD₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
GCD₃(s₁(x), s₁(y), z) -> GCD₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)

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

POL(-₂(x₁, x₂)) = x₁
POL(0) = 0
POL(GCD₃(x₁, x₂, x₃)) = 3·x₁ + 2·x₂ + x₃
POL(max₂(x₁, x₂)) = x₁ + x₂
POL(min₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + 3·x₁

The following usable rules [14] were oriented:

-₂(x, 0) -> x
min₂(0, y) -> 0
max₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

min₂(x, 0) -> 0
min₂(0, y) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(x, 0) -> x
max₂(0, y) -> y
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
gcd₃(s₁(x), s₁(y), z) -> gcd₃(-₂(max₂(x, y), min₂(x, y)), s₁(min₂(x, y)), z)
gcd₃(x, s₁(y), s₁(z)) -> gcd₃(x, -₂(max₂(y, z), min₂(y, z)), s₁(min₂(y, z)))
gcd₃(s₁(x), y, s₁(z)) -> gcd₃(-₂(max₂(x, z), min₂(x, z)), y, s₁(min₂(x, z)))
gcd₃(x, 0, 0) -> x
gcd₃(0, y, 0) -> y
gcd₃(0, 0, z) -> z

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.