Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/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))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 2 SCCs.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

SUM1₁(s₁(x)) -> SUM1₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SUM1₁(x1) = SUM1₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > SUM1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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

                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

SUM₁(s₁(x)) -> SUM₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SUM₁(x1) = SUM₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > SUM₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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.