Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/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))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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 remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
ACK_IN₂(x1, x2) = ACK_IN₁(x1)
s₁(x1) = s₁(x1)
0 = 0
U21₂(x1, x2) = U21₁(x2)
ack_out₁(x1) = x1
ack_in₂(x1, x2) = x1
u22₁(x1) = u22₁(x1)
u21₂(x1, x2) = u21
u11₁(x1) = u11

Lexicographic Path Order [19].
Precedence:

[s₁, U21_1] > ACK_IN₁ > [u21, u11]
[s₁, U21_1] > 0 > [u21, u11]
u22₁ > [u21, u11]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACK_IN₂(x1, x2) = ACK_IN₁(x2)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[ACK_IN₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ 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.