Termination w.r.t. Q proof of TRS/AG01#3.53b.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

Q is empty.

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:

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

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

↳ 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:

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

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

↳ 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:

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

Q is empty.
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) = x3
s(x1) = s(x1)
0 = 0
1 = 1

Recursive path order with status [2].
Quasi-Precedence:

[s₁, 1] > 0

Status:

1: multiset
s₁: [1]
0: multiset

The following usable rules [14] were oriented: none

↳ 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:

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

Q is empty.
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.