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

ackin₂(0, X) -> ackout₁(s₁(X))
ackin₂(s₁(X), 0) -> u11₁(ackin₂(X, s₁(0)))
u11₁(ackout₁(X)) -> ackout₁(X)
ackin₂(s₁(X), s₁(Y)) -> u21₂(ackin₂(s₁(X), Y), X)
u21₂(ackout₁(X), Y) -> u22₁(ackin₂(Y, X))
u22₁(ackout₁(X)) -> ackout₁(X)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

ackin₂(0, X) -> ackout₁(s₁(X))
ackin₂(s₁(X), 0) -> u11₁(ackin₂(X, s₁(0)))
u11₁(ackout₁(X)) -> ackout₁(X)
ackin₂(s₁(X), s₁(Y)) -> u21₂(ackin₂(s₁(X), Y), X)
u21₂(ackout₁(X), Y) -> u22₁(ackin₂(Y, X))
u22₁(ackout₁(X)) -> ackout₁(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:

ackin₂(0, X) -> ackout₁(s₁(X))
ackin₂(s₁(X), 0) -> u11₁(ackin₂(X, s₁(0)))
u11₁(ackout₁(X)) -> ackout₁(X)
ackin₂(s₁(X), s₁(Y)) -> u21₂(ackin₂(s₁(X), Y), X)
u21₂(ackout₁(X), Y) -> u22₁(ackin₂(Y, X))
u22₁(ackout₁(X)) -> ackout₁(X)

The set Q consists of the following terms:

ackin₂(0, x0)
ackin₂(s₁(x0), 0)
u11₁(ackout₁(x0))
ackin₂(s₁(x0), s₁(x1))
u21₂(ackout₁(x0), x1)
u22₁(ackout₁(x0))

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

U21₂(ackout₁(X), Y) -> ACKIN₂(Y, X)
ACKIN₂(s₁(X), s₁(Y)) -> ACKIN₂(s₁(X), Y)
ACKIN₂(s₁(X), 0) -> U11₁(ackin₂(X, s₁(0)))
ACKIN₂(s₁(X), s₁(Y)) -> U21₂(ackin₂(s₁(X), Y), X)
ACKIN₂(s₁(X), 0) -> ACKIN₂(X, s₁(0))
U21₂(ackout₁(X), Y) -> U22₁(ackin₂(Y, X))

The TRS R consists of the following rules:

ackin₂(0, X) -> ackout₁(s₁(X))
ackin₂(s₁(X), 0) -> u11₁(ackin₂(X, s₁(0)))
u11₁(ackout₁(X)) -> ackout₁(X)
ackin₂(s₁(X), s₁(Y)) -> u21₂(ackin₂(s₁(X), Y), X)
u21₂(ackout₁(X), Y) -> u22₁(ackin₂(Y, X))
u22₁(ackout₁(X)) -> ackout₁(X)

The set Q consists of the following terms:

ackin₂(0, x0)
ackin₂(s₁(x0), 0)
u11₁(ackout₁(x0))
ackin₂(s₁(x0), s₁(x1))
u21₂(ackout₁(x0), x1)
u22₁(ackout₁(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:

U21₂(ackout₁(X), Y) -> ACKIN₂(Y, X)
ACKIN₂(s₁(X), s₁(Y)) -> ACKIN₂(s₁(X), Y)
ACKIN₂(s₁(X), 0) -> U11₁(ackin₂(X, s₁(0)))
ACKIN₂(s₁(X), s₁(Y)) -> U21₂(ackin₂(s₁(X), Y), X)
ACKIN₂(s₁(X), 0) -> ACKIN₂(X, s₁(0))
U21₂(ackout₁(X), Y) -> U22₁(ackin₂(Y, X))

The TRS R consists of the following rules:

ackin₂(0, X) -> ackout₁(s₁(X))
ackin₂(s₁(X), 0) -> u11₁(ackin₂(X, s₁(0)))
u11₁(ackout₁(X)) -> ackout₁(X)
ackin₂(s₁(X), s₁(Y)) -> u21₂(ackin₂(s₁(X), Y), X)
u21₂(ackout₁(X), Y) -> u22₁(ackin₂(Y, X))
u22₁(ackout₁(X)) -> ackout₁(X)

The set Q consists of the following terms:

ackin₂(0, x0)
ackin₂(s₁(x0), 0)
u11₁(ackout₁(x0))
ackin₂(s₁(x0), s₁(x1))
u21₂(ackout₁(x0), x1)
u22₁(ackout₁(x0))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

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

ACKIN₂(s₁(X), s₁(Y)) -> ACKIN₂(s₁(X), Y)
U21₂(ackout₁(X), Y) -> ACKIN₂(Y, X)
ACKIN₂(s₁(X), s₁(Y)) -> U21₂(ackin₂(s₁(X), Y), X)
ACKIN₂(s₁(X), 0) -> ACKIN₂(X, s₁(0))

The TRS R consists of the following rules:

ackin₂(0, X) -> ackout₁(s₁(X))
ackin₂(s₁(X), 0) -> u11₁(ackin₂(X, s₁(0)))
u11₁(ackout₁(X)) -> ackout₁(X)
ackin₂(s₁(X), s₁(Y)) -> u21₂(ackin₂(s₁(X), Y), X)
u21₂(ackout₁(X), Y) -> u22₁(ackin₂(Y, X))
u22₁(ackout₁(X)) -> ackout₁(X)

The set Q consists of the following terms:

ackin₂(0, x0)
ackin₂(s₁(x0), 0)
u11₁(ackout₁(x0))
ackin₂(s₁(x0), s₁(x1))
u21₂(ackout₁(x0), x1)
u22₁(ackout₁(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.

ACKIN₂(s₁(X), s₁(Y)) -> U21₂(ackin₂(s₁(X), Y), X)
ACKIN₂(s₁(X), 0) -> ACKIN₂(X, s₁(0))

Used argument filtering: ACKIN₂(x1, x2) = x1
s₁(x1) = s₁(x1)
U21₂(x1, x2) = x2
ackin₂(x1, x2) = ackin
0 = 0
u11₁(x1) = u11
u21₂(x1, x2) = u21
ackout₁(x1) = ackout
u22₁(x1) = u22
Used ordering: Precedence:

ackin > u11 > ackout
ackin > u21 > u22 > ackout

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

U21₂(ackout₁(X), Y) -> ACKIN₂(Y, X)
ACKIN₂(s₁(X), s₁(Y)) -> ACKIN₂(s₁(X), Y)

The TRS R consists of the following rules:

ackin₂(0, X) -> ackout₁(s₁(X))
ackin₂(s₁(X), 0) -> u11₁(ackin₂(X, s₁(0)))
u11₁(ackout₁(X)) -> ackout₁(X)
ackin₂(s₁(X), s₁(Y)) -> u21₂(ackin₂(s₁(X), Y), X)
u21₂(ackout₁(X), Y) -> u22₁(ackin₂(Y, X))
u22₁(ackout₁(X)) -> ackout₁(X)

The set Q consists of the following terms:

ackin₂(0, x0)
ackin₂(s₁(x0), 0)
u11₁(ackout₁(x0))
ackin₂(s₁(x0), s₁(x1))
u21₂(ackout₁(x0), x1)
u22₁(ackout₁(x0))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

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

ACKIN₂(s₁(X), s₁(Y)) -> ACKIN₂(s₁(X), Y)

The TRS R consists of the following rules:

ackin₂(0, X) -> ackout₁(s₁(X))
ackin₂(s₁(X), 0) -> u11₁(ackin₂(X, s₁(0)))
u11₁(ackout₁(X)) -> ackout₁(X)
ackin₂(s₁(X), s₁(Y)) -> u21₂(ackin₂(s₁(X), Y), X)
u21₂(ackout₁(X), Y) -> u22₁(ackin₂(Y, X))
u22₁(ackout₁(X)) -> ackout₁(X)

The set Q consists of the following terms:

ackin₂(0, x0)
ackin₂(s₁(x0), 0)
u11₁(ackout₁(x0))
ackin₂(s₁(x0), s₁(x1))
u21₂(ackout₁(x0), x1)
u22₁(ackout₁(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.

ACKIN₂(s₁(X), s₁(Y)) -> ACKIN₂(s₁(X), Y)

Used argument filtering: ACKIN₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

                        ↳ QDP

                          ↳ PisEmptyProof

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

ackin₂(0, X) -> ackout₁(s₁(X))
ackin₂(s₁(X), 0) -> u11₁(ackin₂(X, s₁(0)))
u11₁(ackout₁(X)) -> ackout₁(X)
ackin₂(s₁(X), s₁(Y)) -> u21₂(ackin₂(s₁(X), Y), X)
u21₂(ackout₁(X), Y) -> u22₁(ackin₂(Y, X))
u22₁(ackout₁(X)) -> ackout₁(X)

The set Q consists of the following terms:

ackin₂(0, x0)
ackin₂(s₁(x0), 0)
u11₁(ackout₁(x0))
ackin₂(s₁(x0), s₁(x1))
u21₂(ackout₁(x0), x1)
u22₁(ackout₁(x0))

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