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:

g₂(x, y) -> x
g₂(x, y) -> y
f₃(0, 1, x) -> f₃(s₁(x), x, x)
f₃(x, y, s₁(z)) -> s₁(f₃(0, 1, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g₂(x, y) -> x
g₂(x, y) -> y
f₃(0, 1, x) -> f₃(s₁(x), x, x)
f₃(x, y, s₁(z)) -> s₁(f₃(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:

F₃(x, y, s₁(z)) -> F₃(0, 1, z)
F₃(0, 1, x) -> F₃(s₁(x), x, x)

The TRS R consists of the following rules:

g₂(x, y) -> x
g₂(x, y) -> y
f₃(0, 1, x) -> f₃(s₁(x), x, x)
f₃(x, y, s₁(z)) -> s₁(f₃(0, 1, z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

F₃(x, y, s₁(z)) -> F₃(0, 1, z)
F₃(0, 1, x) -> F₃(s₁(x), x, x)

The TRS R consists of the following rules:

g₂(x, y) -> x
g₂(x, y) -> y
f₃(0, 1, x) -> f₃(s₁(x), x, x)
f₃(x, y, s₁(z)) -> s₁(f₃(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.

F₃(x, y, s₁(z)) -> F₃(0, 1, z)

The remaining pairs can at least be oriented weakly.

F₃(0, 1, x) -> F₃(s₁(x), x, x)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 1 ) = max{0, -1}

POL( 0 ) = 0

POL( s₁(x₁) ) = 2x₁ + 1

POL( F₃(x₁, ..., x₃) ) = x₃ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F₃(0, 1, x) -> F₃(s₁(x), x, x)

The TRS R consists of the following rules:

g₂(x, y) -> x
g₂(x, y) -> y
f₃(0, 1, x) -> f₃(s₁(x), x, x)
f₃(x, y, s₁(z)) -> s₁(f₃(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.