Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/SK90SLASH2.51.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:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))

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:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))

The set Q consists of the following terms:

ack₂(0, x0)
ack₂(s₁(x0), 0)
ack₂(s₁(x0), s₁(x1))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))

The set Q consists of the following terms:

ack₂(0, x0)
ack₂(s₁(x0), 0)
ack₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

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

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))

The set Q consists of the following terms:

ack₂(0, x0)
ack₂(s₁(x0), 0)
ack₂(s₁(x0), s₁(x1))

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.

ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The remaining pairs can at least by weakly be oriented.

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)

Used ordering: Combined order from the following AFS and order.
ACK₂(x1, x2) = x1
s₁(x1) = s₁(x1)
ack₂(x1, x2) = ack₂(x1, x2)
0 = 0

Lexicographic Path Order [19].
Precedence:

ack₂ > s₁

The following usable rules [14] were oriented:

ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
ack₂(0, y) -> s₁(y)

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)

The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))

The set Q consists of the following terms:

ack₂(0, x0)
ack₂(s₁(x0), 0)
ack₂(s₁(x0), s₁(x1))

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.

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACK₂(x1, x2) = ACK₁(x2)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > ACK₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))

The set Q consists of the following terms:

ack₂(0, x0)
ack₂(s₁(x0), 0)
ack₂(s₁(x0), s₁(x1))

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