Termination w.r.t. Q proof of /tmp/tmpcJ3-yk/tpa06.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(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(x), min(s(y), s(z)))
F(s(x), s(y), s(z)) → P(min(s(x), max(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(y), s(z))
F(s(x), s(y), s(z)) → MIN(s(x), max(s(y), s(z)))
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
F(x, y, 0) → MAX(x, y)
F(x, 0, z) → MAX(x, z)
F(0, y, z) → MAX(y, z)
F(s(x), s(y), s(z)) → MAX(s(y), s(z))
F(s(x), s(y), s(z)) → MAX(s(x), max(s(y), s(z)))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(x), min(s(y), s(z)))
F(s(x), s(y), s(z)) → P(min(s(x), max(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(y), s(z))
F(s(x), s(y), s(z)) → MIN(s(x), max(s(y), s(z)))
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
F(x, y, 0) → MAX(x, y)
F(x, 0, z) → MAX(x, z)
F(0, y, z) → MAX(y, z)
F(s(x), s(y), s(z)) → MAX(s(y), s(z))
F(s(x), s(y), s(z)) → MAX(s(x), max(s(y), s(z)))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 9 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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)

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:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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.

f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

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

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z)))) at position [0,1] we obtained the following new rules:

F(s(x), s(y), s(z)) → F(max(s(x), s(max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

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

F(s(x), s(y), s(z)) → F(max(s(x), s(max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z)))) at position [1,0,1] we obtained the following new rules:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), s(max(y, z)))), min(s(x), min(s(y), s(z))))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), s(max(y, z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(min(s(x), s(max(y, z)))), min(s(x), min(s(y), s(z)))) at position [1,0] we obtained the following new rules:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(s(min(x, max(y, z)))), min(s(x), min(s(y), s(z))))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(s(min(x, max(y, z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), p(s(min(x, max(y, z)))), min(s(x), min(s(y), s(z)))) at position [1] we obtained the following new rules:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z))))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ UsableRulesProof

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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.

p(s(x0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

                                                  ↳ QDP

                                                    ↳ Rewriting

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), min(s(y), s(z)))) at position [2,1] we obtained the following new rules:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), s(min(y, z))))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), s(min(y, z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), min(s(x), s(min(y, z)))) at position [2] we obtained the following new rules:

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), s(min(x, min(y, z))))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ QDPOrderProof

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), s(min(x, min(y, z))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

F(s(x), s(y), s(z)) → F(s(max(x, max(y, z))), min(x, max(y, z)), s(min(x, min(y, z))))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0
POL(F(x₁, x₂, x₃)) = x₂ + x₃
POL(max(x₁, x₂)) = max(x₁, x₂)
POL(min(x₁, x₂)) = min(x₁, x₂)
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ QDPOrderProof

                                                              ↳ QDP

                                                                ↳ PisEmptyProof

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

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

The set Q consists of the following terms:

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