Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/AG01SLASHHASH3.53b.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.

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

      ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₃(x, y, s₁(z)) -> F₃(0, 1, z)

Used argument filtering: F₃(x1, x2, x3) = x3
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ 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 contains 0 SCCs with 1 less node.