Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/variousSLASH22.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(x, 0) -> s₁(0)
f₂(s₁(x), s₁(y)) -> s₁(f₂(x, y))
g₂(0, x) -> g₂(f₂(x, x), x)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₂(x, 0) -> s₁(0)
f₂(s₁(x), s₁(y)) -> s₁(f₂(x, y))
g₂(0, x) -> g₂(f₂(x, x), x)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

f₂(x, 0) -> s₁(0)
f₂(s₁(x), s₁(y)) -> s₁(f₂(x, y))
g₂(0, x) -> g₂(f₂(x, x), x)

The set Q consists of the following terms:

f₂(x0, 0)
f₂(s₁(x0), s₁(x1))
g₂(0, x0)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G₂(0, x) -> F₂(x, x)
G₂(0, x) -> G₂(f₂(x, x), x)
F₂(s₁(x), s₁(y)) -> F₂(x, y)

The TRS R consists of the following rules:

f₂(x, 0) -> s₁(0)
f₂(s₁(x), s₁(y)) -> s₁(f₂(x, y))
g₂(0, x) -> g₂(f₂(x, x), x)

The set Q consists of the following terms:

f₂(x0, 0)
f₂(s₁(x0), s₁(x1))
g₂(0, x0)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

G₂(0, x) -> F₂(x, x)
G₂(0, x) -> G₂(f₂(x, x), x)
F₂(s₁(x), s₁(y)) -> F₂(x, y)

The TRS R consists of the following rules:

f₂(x, 0) -> s₁(0)
f₂(s₁(x), s₁(y)) -> s₁(f₂(x, y))
g₂(0, x) -> g₂(f₂(x, x), x)

The set Q consists of the following terms:

f₂(x0, 0)
f₂(s₁(x0), s₁(x1))
g₂(0, x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

F₂(s₁(x), s₁(y)) -> F₂(x, y)

The TRS R consists of the following rules:

f₂(x, 0) -> s₁(0)
f₂(s₁(x), s₁(y)) -> s₁(f₂(x, y))
g₂(0, x) -> g₂(f₂(x, x), x)

The set Q consists of the following terms:

f₂(x0, 0)
f₂(s₁(x0), s₁(x1))
g₂(0, x0)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

F₂(s₁(x), s₁(y)) -> F₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F₂(x1, x2) = x2
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

f₂(x, 0) -> s₁(0)
f₂(s₁(x), s₁(y)) -> s₁(f₂(x, y))
g₂(0, x) -> g₂(f₂(x, x), x)

The set Q consists of the following terms:

f₂(x0, 0)
f₂(s₁(x0), s₁(x1))
g₂(0, x0)

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