Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/SK90SLASH2.17.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:

sum₁(0) -> 0
sum₁(s₁(x)) -> +₂(sum₁(x), s₁(x))
sum1₁(0) -> 0
sum1₁(s₁(x)) -> s₁(+₂(sum1₁(x), +₂(x, x)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

sum₁(0) -> 0
sum₁(s₁(x)) -> +₂(sum₁(x), s₁(x))
sum1₁(0) -> 0
sum1₁(s₁(x)) -> s₁(+₂(sum1₁(x), +₂(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:

sum₁(0) -> 0
sum₁(s₁(x)) -> +₂(sum₁(x), s₁(x))
sum1₁(0) -> 0
sum1₁(s₁(x)) -> s₁(+₂(sum1₁(x), +₂(x, x)))

The set Q consists of the following terms:

sum₁(0)
sum₁(s₁(x0))
sum1₁(0)
sum1₁(s₁(x0))

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

SUM1₁(s₁(x)) -> SUM1₁(x)
SUM₁(s₁(x)) -> SUM₁(x)

The TRS R consists of the following rules:

sum₁(0) -> 0
sum₁(s₁(x)) -> +₂(sum₁(x), s₁(x))
sum1₁(0) -> 0
sum1₁(s₁(x)) -> s₁(+₂(sum1₁(x), +₂(x, x)))

The set Q consists of the following terms:

sum₁(0)
sum₁(s₁(x0))
sum1₁(0)
sum1₁(s₁(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:

SUM1₁(s₁(x)) -> SUM1₁(x)
SUM₁(s₁(x)) -> SUM₁(x)

The TRS R consists of the following rules:

sum₁(0) -> 0
sum₁(s₁(x)) -> +₂(sum₁(x), s₁(x))
sum1₁(0) -> 0
sum1₁(s₁(x)) -> s₁(+₂(sum1₁(x), +₂(x, x)))

The set Q consists of the following terms:

sum₁(0)
sum₁(s₁(x0))
sum1₁(0)
sum1₁(s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

SUM1₁(s₁(x)) -> SUM1₁(x)

The TRS R consists of the following rules:

sum₁(0) -> 0
sum₁(s₁(x)) -> +₂(sum₁(x), s₁(x))
sum1₁(0) -> 0
sum1₁(s₁(x)) -> s₁(+₂(sum1₁(x), +₂(x, x)))

The set Q consists of the following terms:

sum₁(0)
sum₁(s₁(x0))
sum1₁(0)
sum1₁(s₁(x0))

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.

SUM1₁(s₁(x)) -> SUM1₁(x)

Used argument filtering: SUM1₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

sum₁(0) -> 0
sum₁(s₁(x)) -> +₂(sum₁(x), s₁(x))
sum1₁(0) -> 0
sum1₁(s₁(x)) -> s₁(+₂(sum1₁(x), +₂(x, x)))

The set Q consists of the following terms:

sum₁(0)
sum₁(s₁(x0))
sum1₁(0)
sum1₁(s₁(x0))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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

SUM₁(s₁(x)) -> SUM₁(x)

The TRS R consists of the following rules:

sum₁(0) -> 0
sum₁(s₁(x)) -> +₂(sum₁(x), s₁(x))
sum1₁(0) -> 0
sum1₁(s₁(x)) -> s₁(+₂(sum1₁(x), +₂(x, x)))

The set Q consists of the following terms:

sum₁(0)
sum₁(s₁(x0))
sum1₁(0)
sum1₁(s₁(x0))

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.

SUM₁(s₁(x)) -> SUM₁(x)

Used argument filtering: SUM₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

sum₁(0) -> 0
sum₁(s₁(x)) -> +₂(sum₁(x), s₁(x))
sum1₁(0) -> 0
sum1₁(s₁(x)) -> s₁(+₂(sum1₁(x), +₂(x, x)))

The set Q consists of the following terms:

sum₁(0)
sum₁(s₁(x0))
sum1₁(0)
sum1₁(s₁(x0))

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