Termination w.r.t. Q proof of /tmp/tmpYwRuPb/tpa1.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:

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

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

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)
min(x, 0) → 0

The TRS R 2 is

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

The signature Sigma is {gcd}

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 3 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

R is empty.
The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

-¹(s(x), s(y)) → -¹(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

MAX(s(x), s(y)) → MAX(x, y)

R is empty.
The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

MAX(s(x), s(y)) → MAX(x, y)

From the DPs we obtained the following set of size-change graphs:

MAX(s(x), s(y)) → MAX(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

MIN(s(x), s(y)) → MIN(x, y)

R is empty.
The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

MIN(s(x), s(y)) → MIN(x, y)

From the DPs we obtained the following set of size-change graphs:

MIN(s(x), s(y)) → MIN(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

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

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

gcd(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y))) at position [0] we obtained the following new rules:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( -(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	0	0	\
\	0	0	/

x₂

M( max(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	1	0	\
\	0	1	/

x₂

M( s(x₁) ) =

/	1	\
\	0	/

/	1	1	\
\	1	1	/

x₁

M( min(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	0	0	\
\	0	0	/

x₂

M( 0 ) =

/	0	\
\	0	/

Tuple symbols:

M( GCD(x₁, x₂) ) =

[

]

x₁

[

]

x₂

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

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

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

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.