Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/nonterminSLASHAG01SLASHHASH4.33.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:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

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:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

The set Q consists of the following terms:

sum₂(cons₂(s₁(x0), x1), cons₂(x2, x3))
sum₂(cons₂(0, x0), x1)
sum₂(nil, x0)
weight₁(cons₂(x0, cons₂(x1, x2)))
weight₁(cons₂(x0, nil))

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

WEIGHT₁(cons₂(n, cons₂(m, x))) -> SUM₂(cons₂(n, cons₂(m, x)), cons₂(0, x))
WEIGHT₁(cons₂(n, cons₂(m, x))) -> WEIGHT₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
SUM₂(cons₂(0, x), y) -> SUM₂(x, y)
SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

The set Q consists of the following terms:

sum₂(cons₂(s₁(x0), x1), cons₂(x2, x3))
sum₂(cons₂(0, x0), x1)
sum₂(nil, x0)
weight₁(cons₂(x0, cons₂(x1, x2)))
weight₁(cons₂(x0, nil))

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:

WEIGHT₁(cons₂(n, cons₂(m, x))) -> SUM₂(cons₂(n, cons₂(m, x)), cons₂(0, x))
WEIGHT₁(cons₂(n, cons₂(m, x))) -> WEIGHT₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
SUM₂(cons₂(0, x), y) -> SUM₂(x, y)
SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

The set Q consists of the following terms:

sum₂(cons₂(s₁(x0), x1), cons₂(x2, x3))
sum₂(cons₂(0, x0), x1)
sum₂(nil, x0)
weight₁(cons₂(x0, cons₂(x1, x2)))
weight₁(cons₂(x0, nil))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))
SUM₂(cons₂(0, x), y) -> SUM₂(x, y)

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

The set Q consists of the following terms:

sum₂(cons₂(s₁(x0), x1), cons₂(x2, x3))
sum₂(cons₂(0, x0), x1)
sum₂(nil, x0)
weight₁(cons₂(x0, cons₂(x1, x2)))
weight₁(cons₂(x0, nil))

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.

SUM₂(cons₂(0, x), y) -> SUM₂(x, y)

Used argument filtering: SUM₂(x1, x2) = x1
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

              ↳ QDP

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

SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

The set Q consists of the following terms:

sum₂(cons₂(s₁(x0), x1), cons₂(x2, x3))
sum₂(cons₂(0, x0), x1)
sum₂(nil, x0)
weight₁(cons₂(x0, cons₂(x1, x2)))
weight₁(cons₂(x0, nil))

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.

SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))

Used argument filtering: SUM₂(x1, x2) = x1
cons₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

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

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

The set Q consists of the following terms:

sum₂(cons₂(s₁(x0), x1), cons₂(x2, x3))
sum₂(cons₂(0, x0), x1)
sum₂(nil, x0)
weight₁(cons₂(x0, cons₂(x1, x2)))
weight₁(cons₂(x0, nil))

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:

WEIGHT₁(cons₂(n, cons₂(m, x))) -> WEIGHT₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

The set Q consists of the following terms:

sum₂(cons₂(s₁(x0), x1), cons₂(x2, x3))
sum₂(cons₂(0, x0), x1)
sum₂(nil, x0)
weight₁(cons₂(x0, cons₂(x1, x2)))
weight₁(cons₂(x0, nil))

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.

WEIGHT₁(cons₂(n, cons₂(m, x))) -> WEIGHT₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))

Used argument filtering: WEIGHT₁(x1) = x1
cons₂(x1, x2) = cons₁(x2)
sum₂(x1, x2) = x2
Used ordering: 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:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

The set Q consists of the following terms:

sum₂(cons₂(s₁(x0), x1), cons₂(x2, x3))
sum₂(cons₂(0, x0), x1)
sum₂(nil, x0)
weight₁(cons₂(x0, cons₂(x1, x2)))
weight₁(cons₂(x0, nil))

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