Termination w.r.t. Q proof of TRS/AG01#3.52.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(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, 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(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

The set Q consists of the following terms:

f(0, 1, x0)
f(x0, x1, s(x2))

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(x, y, s(z)) → F(0, 1, z)
F(0, 1, x) → F(s(x), x, x)

The TRS R consists of the following rules:

f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

The set Q consists of the following terms:

f(0, 1, x0)
f(x0, x1, s(x2))

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(x, y, s(z)) → F(0, 1, z)
F(0, 1, x) → F(s(x), x, x)

The TRS R consists of the following rules:

f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

The set Q consists of the following terms:

f(0, 1, x0)
f(x0, x1, s(x2))

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

              ↳ QDPOrderProof

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

F(x, y, s(z)) → F(0, 1, z)
F(0, 1, x) → F(s(x), x, x)

The TRS R consists of the following rules:

f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

The set Q consists of the following terms:

f(0, 1, x0)
f(x0, x1, s(x2))

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, y, s(z)) → F(0, 1, z)

The remaining pairs can at least be oriented weakly.

F(0, 1, x) → F(s(x), x, x)

Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3) = F(x3)
s(x1) = s(x1)
0 = 0
1 = 1

Recursive Path Order [2].
Precedence:

0 > [F₁, s_1]
1 > [F₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

F(0, 1, x) → F(s(x), x, x)

The TRS R consists of the following rules:

f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

The set Q consists of the following terms:

f(0, 1, x0)
f(x0, x1, s(x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.