Termination w.r.t. Q proof of TRS/SK902.52.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:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(s(0), y, z) → F(0, s(y), s(z))
F(s(x), 0, s(z)) → F(x, s(0), z)
F(s(x), s(y), 0) → F(x, y, s(0))
F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))
F(s(x), s(y), s(z)) → F(s(x), s(y), z)
F(0, s(s(y)), s(s(z))) → F(0, y, f(0, s(s(y)), s(z)))
F(0, s(0), s(s(z))) → F(0, s(0), z)
F(0, s(s(y)), s(0)) → F(0, y, s(0))
F(s(x), s(y), s(z)) → F(x, y, f(s(x), s(y), z))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

F(s(0), y, z) → F(0, s(y), s(z))
F(s(x), 0, s(z)) → F(x, s(0), z)
F(s(x), s(y), 0) → F(x, y, s(0))
F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))
F(s(x), s(y), s(z)) → F(s(x), s(y), z)
F(0, s(s(y)), s(s(z))) → F(0, y, f(0, s(s(y)), s(z)))
F(0, s(0), s(s(z))) → F(0, s(0), z)
F(0, s(s(y)), s(0)) → F(0, y, s(0))
F(s(x), s(y), s(z)) → F(x, y, f(s(x), s(y), z))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

F(s(0), y, z) → F(0, s(y), s(z))
F(s(x), s(y), 0) → F(x, y, s(0))
F(s(x), 0, s(z)) → F(x, s(0), z)
F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))
F(s(x), s(y), s(z)) → F(s(x), s(y), z)
F(0, s(s(y)), s(0)) → F(0, y, s(0))
F(0, s(0), s(s(z))) → F(0, s(0), z)
F(0, s(s(y)), s(s(z))) → F(0, y, f(0, s(s(y)), s(z)))
F(s(x), s(y), s(z)) → F(x, y, f(s(x), s(y), z))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

F(0, s(0), s(s(z))) → F(0, s(0), z)

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

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.

F(0, s(0), s(s(z))) → F(0, s(0), z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3) = F(x1, x3)
0 = 0
s(x1) = s(x1)

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

[F₂, s_1] > 0

Status:

0: multiset
s₁: [1]
F₂: [1,2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

F(0, s(s(y)), s(0)) → F(0, y, s(0))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

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.

F(0, s(s(y)), s(0)) → F(0, y, s(0))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3) = F(x1, x2, x3)
0 = 0
s(x1) = s(x1)

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

0 > F₃ > s₁

Status:

0: multiset
s₁: [1]
F₃: [1,3,2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))
F(0, s(s(y)), s(s(z))) → F(0, y, f(0, s(s(y)), s(z)))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

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.

F(0, s(s(y)), s(s(z))) → F(0, y, f(0, s(s(y)), s(z)))

The remaining pairs can at least be oriented weakly.

F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))

Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3) = F(x2)
0 = 0
s(x1) = s(x1)
f(x1, x2, x3) = x1

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

[0, s_1] > F₁

Status:

0: multiset
s₁: [1]
F₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                  ↳ QDP

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

F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

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.

F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))

s₁ > 0 > F₂

Status:

0: multiset
s₁: [1]
F₂: [2,1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

                  ↳ QDP

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

F(s(x), s(y), 0) → F(x, y, s(0))
F(s(x), 0, s(z)) → F(x, s(0), z)
F(s(x), s(y), s(z)) → F(s(x), s(y), z)
F(s(x), s(y), s(z)) → F(x, y, f(s(x), s(y), z))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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

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.

F(s(x), s(y), 0) → F(x, y, s(0))
F(s(x), 0, s(z)) → F(x, s(0), z)
F(s(x), s(y), s(z)) → F(s(x), s(y), z)
F(s(x), s(y), s(z)) → F(x, y, f(s(x), s(y), z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3) = F(x1, x2, x3)
s(x1) = s(x1)
0 = 0
f(x1, x2, x3) = f(x1, x2, x3)

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

F₃ > f₃ > 0 > s₁

Status:

f₃: [1,2,3]
0: multiset
s₁: [1]
F₃: [1,3,2]

The following usable rules [14] were oriented:

f(0, y, 0) → s(y)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(s(0), y, z) → f(0, s(y), s(z))
f(x, 0, 0) → s(x)
f(0, 0, z) → s(z)
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

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