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

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:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(0, 0, z) → z

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(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:

MIN(s(x), s(y)) → MIN(x, y)
-¹(s(x), s(y)) → -¹(x, y)
GCD(x, s(y), s(z)) → GCD(x, -(max(y, z), min(y, z)), s(min(y, z)))
GCD(s(x), s(y), z) → -¹(max(x, y), min(x, y))
GCD(s(x), y, s(z)) → GCD(-(max(x, z), min(x, z)), y, s(min(x, z)))
GCD(x, s(y), s(z)) → -¹(max(y, z), min(y, z))
GCD(s(x), s(y), z) → GCD(-(max(x, y), min(x, y)), s(min(x, y)), z)
GCD(x, s(y), s(z)) → MIN(y, z)
GCD(s(x), y, s(z)) → MAX(x, z)
GCD(x, s(y), s(z)) → MAX(y, z)
GCD(s(x), s(y), z) → MIN(x, y)
GCD(s(x), s(y), z) → MAX(x, y)
MAX(s(x), s(y)) → MAX(x, y)
GCD(s(x), y, s(z)) → MIN(x, z)
GCD(s(x), y, s(z)) → -¹(max(x, z), min(x, z))

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(0, 0, z) → z

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

MIN(s(x), s(y)) → MIN(x, y)
-¹(s(x), s(y)) → -¹(x, y)
GCD(x, s(y), s(z)) → GCD(x, -(max(y, z), min(y, z)), s(min(y, z)))
GCD(s(x), s(y), z) → -¹(max(x, y), min(x, y))
GCD(s(x), y, s(z)) → GCD(-(max(x, z), min(x, z)), y, s(min(x, z)))
GCD(x, s(y), s(z)) → -¹(max(y, z), min(y, z))
GCD(s(x), s(y), z) → GCD(-(max(x, y), min(x, y)), s(min(x, y)), z)
GCD(x, s(y), s(z)) → MIN(y, z)
GCD(s(x), y, s(z)) → MAX(x, z)
GCD(x, s(y), s(z)) → MAX(y, z)
GCD(s(x), s(y), z) → MIN(x, y)
GCD(s(x), s(y), z) → MAX(x, y)
MAX(s(x), s(y)) → MAX(x, y)
GCD(s(x), y, s(z)) → MIN(x, z)
GCD(s(x), y, s(z)) → -¹(max(x, z), min(x, z))

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(0, 0, z) → z

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MIN(s(x), s(y)) → MIN(x, y)
GCD(x, s(y), s(z)) → GCD(x, -(max(y, z), min(y, z)), s(min(y, z)))
-¹(s(x), s(y)) → -¹(x, y)
GCD(s(x), s(y), z) → -¹(max(x, y), min(x, y))
GCD(s(x), y, s(z)) → GCD(-(max(x, z), min(x, z)), y, s(min(x, z)))
GCD(x, s(y), s(z)) → -¹(max(y, z), min(y, z))
GCD(s(x), s(y), z) → GCD(-(max(x, y), min(x, y)), s(min(x, y)), z)
GCD(x, s(y), s(z)) → MIN(y, z)
GCD(s(x), y, s(z)) → MAX(x, z)
GCD(x, s(y), s(z)) → MAX(y, z)
GCD(s(x), s(y), z) → MIN(x, y)
GCD(s(x), s(y), z) → MAX(x, y)
MAX(s(x), s(y)) → MAX(x, y)
GCD(s(x), y, s(z)) → -¹(max(x, z), min(x, z))
GCD(s(x), y, s(z)) → MIN(x, z)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(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.

-¹(s(x), s(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2) = x2
s(x1) = s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

MAX(s(x), s(y)) → MAX(x, y)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(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.

MAX(s(x), s(y)) → MAX(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MAX(x1, x2) = x2
s(x1) = s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

MIN(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(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.

MIN(s(x), s(y)) → MIN(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MIN(x1, x2) = x2
s(x1) = s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

GCD(x, s(y), s(z)) → GCD(x, -(max(y, z), min(y, z)), s(min(y, z)))
GCD(s(x), y, s(z)) → GCD(-(max(x, z), min(x, z)), y, s(min(x, z)))
GCD(s(x), s(y), z) → GCD(-(max(x, y), min(x, y)), s(min(x, y)), z)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y), z) → gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd(x, s(y), s(z)) → gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z)) → gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))
gcd(x, 0, 0) → x
gcd(0, y, 0) → y
gcd(0, 0, z) → z

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