Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/secret2005SLASHtpa1.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(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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(min₂(x, y)))

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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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:

GCD₂(s₁(x), s₁(y)) -> -¹₂(s₁(max₂(x, y)), s₁(min₂(x, y)))
GCD₂(s₁(x), s₁(y)) -> GCD₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(min₂(x, y)))
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
GCD₂(s₁(x), s₁(y)) -> MAX₂(x, y)
GCD₂(s₁(x), s₁(y)) -> MIN₂(x, y)
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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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:

GCD₂(s₁(x), s₁(y)) -> -¹₂(s₁(max₂(x, y)), s₁(min₂(x, y)))
GCD₂(s₁(x), s₁(y)) -> GCD₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(min₂(x, y)))
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
GCD₂(s₁(x), s₁(y)) -> MAX₂(x, y)
GCD₂(s₁(x), s₁(y)) -> MIN₂(x, y)
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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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 3 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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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 strictly oriented and are deleted.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
-¹₂(x1, x2) = -¹₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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:

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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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 strictly oriented and are deleted.

MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MAX₂(x1, x2) = MAX₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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:

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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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 strictly oriented and are deleted.

MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MIN₂(x1, x2) = MIN₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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:

GCD₂(s₁(x), s₁(y)) -> GCD₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(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)
gcd₂(s₁(x), s₁(y)) -> gcd₂(-₂(s₁(max₂(x, y)), s₁(min₂(x, y))), s₁(min₂(x, y)))

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