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

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:

ack_in₂(0, n) -> ack_out₁(s₁(n))
ack_in₂(s₁(m), 0) -> u11₁(ack_in₂(m, s₁(0)))
u11₁(ack_out₁(n)) -> ack_out₁(n)
ack_in₂(s₁(m), s₁(n)) -> u21₂(ack_in₂(s₁(m), n), m)
u21₂(ack_out₁(n), m) -> u22₁(ack_in₂(m, n))
u22₁(ack_out₁(n)) -> ack_out₁(n)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

ack_in₂(0, n) -> ack_out₁(s₁(n))
ack_in₂(s₁(m), 0) -> u11₁(ack_in₂(m, s₁(0)))
u11₁(ack_out₁(n)) -> ack_out₁(n)
ack_in₂(s₁(m), s₁(n)) -> u21₂(ack_in₂(s₁(m), n), m)
u21₂(ack_out₁(n), m) -> u22₁(ack_in₂(m, n))
u22₁(ack_out₁(n)) -> ack_out₁(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:

ack_in₂(0, n) -> ack_out₁(s₁(n))
ack_in₂(s₁(m), 0) -> u11₁(ack_in₂(m, s₁(0)))
u11₁(ack_out₁(n)) -> ack_out₁(n)
ack_in₂(s₁(m), s₁(n)) -> u21₂(ack_in₂(s₁(m), n), m)
u21₂(ack_out₁(n), m) -> u22₁(ack_in₂(m, n))
u22₁(ack_out₁(n)) -> ack_out₁(n)

The set Q consists of the following terms:

ack_in₂(0, x0)
ack_in₂(s₁(x0), 0)
u11₁(ack_out₁(x0))
ack_in₂(s₁(x0), s₁(x1))
u21₂(ack_out₁(x0), x1)
u22₁(ack_out₁(x0))

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

ACK_IN₂(s₁(m), s₁(n)) -> ACK_IN₂(s₁(m), n)
ACK_IN₂(s₁(m), 0) -> U11₁(ack_in₂(m, s₁(0)))
ACK_IN₂(s₁(m), 0) -> ACK_IN₂(m, s₁(0))
U21₂(ack_out₁(n), m) -> ACK_IN₂(m, n)
ACK_IN₂(s₁(m), s₁(n)) -> U21₂(ack_in₂(s₁(m), n), m)
U21₂(ack_out₁(n), m) -> U22₁(ack_in₂(m, n))

The TRS R consists of the following rules:

ack_in₂(0, n) -> ack_out₁(s₁(n))
ack_in₂(s₁(m), 0) -> u11₁(ack_in₂(m, s₁(0)))
u11₁(ack_out₁(n)) -> ack_out₁(n)
ack_in₂(s₁(m), s₁(n)) -> u21₂(ack_in₂(s₁(m), n), m)
u21₂(ack_out₁(n), m) -> u22₁(ack_in₂(m, n))
u22₁(ack_out₁(n)) -> ack_out₁(n)

The set Q consists of the following terms:

ack_in₂(0, x0)
ack_in₂(s₁(x0), 0)
u11₁(ack_out₁(x0))
ack_in₂(s₁(x0), s₁(x1))
u21₂(ack_out₁(x0), x1)
u22₁(ack_out₁(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:

ACK_IN₂(s₁(m), s₁(n)) -> ACK_IN₂(s₁(m), n)
ACK_IN₂(s₁(m), 0) -> U11₁(ack_in₂(m, s₁(0)))
ACK_IN₂(s₁(m), 0) -> ACK_IN₂(m, s₁(0))
U21₂(ack_out₁(n), m) -> ACK_IN₂(m, n)
ACK_IN₂(s₁(m), s₁(n)) -> U21₂(ack_in₂(s₁(m), n), m)
U21₂(ack_out₁(n), m) -> U22₁(ack_in₂(m, n))

The TRS R consists of the following rules:

ack_in₂(0, n) -> ack_out₁(s₁(n))
ack_in₂(s₁(m), 0) -> u11₁(ack_in₂(m, s₁(0)))
u11₁(ack_out₁(n)) -> ack_out₁(n)
ack_in₂(s₁(m), s₁(n)) -> u21₂(ack_in₂(s₁(m), n), m)
u21₂(ack_out₁(n), m) -> u22₁(ack_in₂(m, n))
u22₁(ack_out₁(n)) -> ack_out₁(n)

The set Q consists of the following terms:

ack_in₂(0, x0)
ack_in₂(s₁(x0), 0)
u11₁(ack_out₁(x0))
ack_in₂(s₁(x0), s₁(x1))
u21₂(ack_out₁(x0), x1)
u22₁(ack_out₁(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:

ACK_IN₂(s₁(m), s₁(n)) -> ACK_IN₂(s₁(m), n)
ACK_IN₂(s₁(m), 0) -> ACK_IN₂(m, s₁(0))
U21₂(ack_out₁(n), m) -> ACK_IN₂(m, n)
ACK_IN₂(s₁(m), s₁(n)) -> U21₂(ack_in₂(s₁(m), n), m)

The TRS R consists of the following rules:

ack_in₂(0, n) -> ack_out₁(s₁(n))
ack_in₂(s₁(m), 0) -> u11₁(ack_in₂(m, s₁(0)))
u11₁(ack_out₁(n)) -> ack_out₁(n)
ack_in₂(s₁(m), s₁(n)) -> u21₂(ack_in₂(s₁(m), n), m)
u21₂(ack_out₁(n), m) -> u22₁(ack_in₂(m, n))
u22₁(ack_out₁(n)) -> ack_out₁(n)

The set Q consists of the following terms:

ack_in₂(0, x0)
ack_in₂(s₁(x0), 0)
u11₁(ack_out₁(x0))
ack_in₂(s₁(x0), s₁(x1))
u21₂(ack_out₁(x0), x1)
u22₁(ack_out₁(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.

ACK_IN₂(s₁(m), 0) -> ACK_IN₂(m, s₁(0))
ACK_IN₂(s₁(m), s₁(n)) -> U21₂(ack_in₂(s₁(m), n), m)

Used argument filtering: ACK_IN₂(x1, x2) = x1
s₁(x1) = s₁(x1)
U21₂(x1, x2) = x2
ack_in₂(x1, x2) = ack_in
0 = 0
u11₁(x1) = u11
u21₂(x1, x2) = u21
ack_out₁(x1) = ack_out
u22₁(x1) = u22
Used ordering: Precedence:

ack_in > u11 > ack_out
ack_in > u21 > u22 > ack_out

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

ACK_IN₂(s₁(m), s₁(n)) -> ACK_IN₂(s₁(m), n)
U21₂(ack_out₁(n), m) -> ACK_IN₂(m, n)

The TRS R consists of the following rules:

ack_in₂(0, n) -> ack_out₁(s₁(n))
ack_in₂(s₁(m), 0) -> u11₁(ack_in₂(m, s₁(0)))
u11₁(ack_out₁(n)) -> ack_out₁(n)
ack_in₂(s₁(m), s₁(n)) -> u21₂(ack_in₂(s₁(m), n), m)
u21₂(ack_out₁(n), m) -> u22₁(ack_in₂(m, n))
u22₁(ack_out₁(n)) -> ack_out₁(n)

The set Q consists of the following terms:

ack_in₂(0, x0)
ack_in₂(s₁(x0), 0)
u11₁(ack_out₁(x0))
ack_in₂(s₁(x0), s₁(x1))
u21₂(ack_out₁(x0), x1)
u22₁(ack_out₁(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:

ACK_IN₂(s₁(m), s₁(n)) -> ACK_IN₂(s₁(m), n)

The TRS R consists of the following rules:

ack_in₂(0, n) -> ack_out₁(s₁(n))
ack_in₂(s₁(m), 0) -> u11₁(ack_in₂(m, s₁(0)))
u11₁(ack_out₁(n)) -> ack_out₁(n)
ack_in₂(s₁(m), s₁(n)) -> u21₂(ack_in₂(s₁(m), n), m)
u21₂(ack_out₁(n), m) -> u22₁(ack_in₂(m, n))
u22₁(ack_out₁(n)) -> ack_out₁(n)

The set Q consists of the following terms:

ack_in₂(0, x0)
ack_in₂(s₁(x0), 0)
u11₁(ack_out₁(x0))
ack_in₂(s₁(x0), s₁(x1))
u21₂(ack_out₁(x0), x1)
u22₁(ack_out₁(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.

ACK_IN₂(s₁(m), s₁(n)) -> ACK_IN₂(s₁(m), n)

Used argument filtering: ACK_IN₂(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:

ack_in₂(0, n) -> ack_out₁(s₁(n))
ack_in₂(s₁(m), 0) -> u11₁(ack_in₂(m, s₁(0)))
u11₁(ack_out₁(n)) -> ack_out₁(n)
ack_in₂(s₁(m), s₁(n)) -> u21₂(ack_in₂(s₁(m), n), m)
u21₂(ack_out₁(n), m) -> u22₁(ack_in₂(m, n))
u22₁(ack_out₁(n)) -> ack_out₁(n)

The set Q consists of the following terms:

ack_in₂(0, x0)
ack_in₂(s₁(x0), 0)
u11₁(ack_out₁(x0))
ack_in₂(s₁(x0), s₁(x1))
u21₂(ack_out₁(x0), x1)
u22₁(ack_out₁(x0))

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