Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/SK90SLASH2.07.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:

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(x, y)) -> y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(x, y)) -> y

Q is empty.

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

F₂(g₂(x, y), z) -> F₂(y, z)
F₂(g₂(x, y), z) -> F₂(x, z)
F₂(f₂(x, y), z) -> F₂(x, f₂(y, z))
F₂(f₂(x, y), z) -> F₂(y, z)

The TRS R consists of the following rules:

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(x, y)) -> y

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:

F₂(g₂(x, y), z) -> F₂(y, z)
F₂(g₂(x, y), z) -> F₂(x, z)
F₂(f₂(x, y), z) -> F₂(x, f₂(y, z))
F₂(f₂(x, y), z) -> F₂(y, z)

The TRS R consists of the following rules:

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(x, y)) -> y

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.

F₂(g₂(x, y), z) -> F₂(y, z)
F₂(g₂(x, y), z) -> F₂(x, z)
F₂(f₂(x, y), z) -> F₂(x, f₂(y, z))
F₂(f₂(x, y), z) -> F₂(y, z)

Used argument filtering: F₂(x1, x2) = x1
g₂(x1, x2) = g₂(x1, x2)
f₂(x1, x2) = f₂(x1, x2)
0 = 0
i₁(x1) = i
1 = 1
2 = 2
Used ordering: Precedence:

f₂ > g₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ PisEmptyProof

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

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(x, y)) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.