Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/SK90SLASH2.28.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:

dfib₂(s₁(s₁(x)), y) -> dfib₂(s₁(x), dfib₂(x, y))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

dfib₂(s₁(s₁(x)), y) -> dfib₂(s₁(x), dfib₂(x, y))

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:

dfib₂(s₁(s₁(x)), y) -> dfib₂(s₁(x), dfib₂(x, y))

The set Q consists of the following terms:

dfib₂(s₁(s₁(x0)), x1)

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

DFIB₂(s₁(s₁(x)), y) -> DFIB₂(x, y)
DFIB₂(s₁(s₁(x)), y) -> DFIB₂(s₁(x), dfib₂(x, y))

The TRS R consists of the following rules:

dfib₂(s₁(s₁(x)), y) -> dfib₂(s₁(x), dfib₂(x, y))

The set Q consists of the following terms:

dfib₂(s₁(s₁(x0)), x1)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

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

DFIB₂(s₁(s₁(x)), y) -> DFIB₂(x, y)
DFIB₂(s₁(s₁(x)), y) -> DFIB₂(s₁(x), dfib₂(x, y))

The TRS R consists of the following rules:

dfib₂(s₁(s₁(x)), y) -> dfib₂(s₁(x), dfib₂(x, y))

The set Q consists of the following terms:

dfib₂(s₁(s₁(x0)), x1)

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.

DFIB₂(s₁(s₁(x)), y) -> DFIB₂(x, y)
DFIB₂(s₁(s₁(x)), y) -> DFIB₂(s₁(x), dfib₂(x, y))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DFIB₂(x1, x2) = x1
s₁(x1) = s₁(x1)
dfib₂(x1, x2) = dfib₁(x2)

Lexicographic Path Order [19].
Precedence:

dfib₁ > s₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

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

dfib₂(s₁(s₁(x)), y) -> dfib₂(s₁(x), dfib₂(x, y))

The set Q consists of the following terms:

dfib₂(s₁(s₁(x0)), x1)

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