Termination w.r.t. Q proof of TRS/nontermin/AG01#4.32.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, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

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, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), 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(x), s(y)) → F(x, s(c(s(y))))
F(x, c(y)) → F(y, y)
F(x, c(y)) → F(x, s(f(y, y)))

The TRS R consists of the following rules:

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), 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(x), s(y)) → F(x, s(c(s(y))))
F(x, c(y)) → F(y, y)
F(x, c(y)) → F(x, s(f(y, y)))

The TRS R consists of the following rules:

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), 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(x), s(y)) → F(x, s(c(s(y))))
F(x, c(y)) → F(y, y)
F(x, c(y)) → F(x, s(f(y, y)))

The TRS R consists of the following rules:

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

F(s(x), s(y)) → F(x, s(c(s(y))))

The TRS R consists of the following rules:

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), 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)) → F(x, s(c(s(y))))

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

Recursive path order with status [2].
Precedence:

trivial

Status:

s₁: multiset
F₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

The set Q consists of the following terms:

f(x0, c(x1))
f(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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

F(x, c(y)) → F(y, y)

The TRS R consists of the following rules:

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), 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(x, c(y)) → F(y, y)

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

Recursive path order with status [2].
Precedence:

c₁ > F₂

Status:

c₁: multiset
F₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

The set Q consists of the following terms:

f(x0, c(x1))
f(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.