Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/SK90SLASH2.21.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:

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(x, y))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(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:

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(x, y))

The set Q consists of the following terms:

bin₂(x0, 0)
bin₂(0, s₁(x0))
bin₂(s₁(x0), s₁(x1))

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

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, y)
BIN₂(s₁(x), s₁(y)) -> BIN₂(x, s₁(y))

The TRS R consists of the following rules:

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(x, y))

The set Q consists of the following terms:

bin₂(x0, 0)
bin₂(0, s₁(x0))
bin₂(s₁(x0), s₁(x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, y)
BIN₂(s₁(x), s₁(y)) -> BIN₂(x, s₁(y))

The TRS R consists of the following rules:

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(x, y))

The set Q consists of the following terms:

bin₂(x0, 0)
bin₂(0, s₁(x0))
bin₂(s₁(x0), s₁(x1))

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.

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, y)

Used argument filtering: BIN₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ QDPAfsSolverProof

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

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, s₁(y))

The TRS R consists of the following rules:

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(x, y))

The set Q consists of the following terms:

bin₂(x0, 0)
bin₂(0, s₁(x0))
bin₂(s₁(x0), s₁(x1))

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.

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, s₁(y))

Used argument filtering: BIN₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ PisEmptyProof

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

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(x, y))

The set Q consists of the following terms:

bin₂(x0, 0)
bin₂(0, s₁(x0))
bin₂(s₁(x0), s₁(x1))

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