Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/AProVESLASHrta1.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

plus₂(s₁(s₁(x)), y) -> s₁(plus₂(x, s₁(y)))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(s₁(0), y) -> s₁(y)
plus₂(0, y) -> y
ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, plus₂(y, ack₂(s₁(x), y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus₂(s₁(s₁(x)), y) -> s₁(plus₂(x, s₁(y)))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(s₁(0), y) -> s₁(y)
plus₂(0, y) -> y
ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, plus₂(y, ack₂(s₁(x), y)))

Q is empty.

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, plus₂(y, ack₂(s₁(x), y)))
PLUS₂(x, s₁(s₁(y))) -> PLUS₂(s₁(x), y)
ACK₂(s₁(x), s₁(y)) -> PLUS₂(y, ack₂(s₁(x), y))
PLUS₂(s₁(s₁(x)), y) -> PLUS₂(x, s₁(y))
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

plus₂(s₁(s₁(x)), y) -> s₁(plus₂(x, s₁(y)))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(s₁(0), y) -> s₁(y)
plus₂(0, y) -> y
ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, plus₂(y, ack₂(s₁(x), y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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, plus₂(y, ack₂(s₁(x), y)))
PLUS₂(x, s₁(s₁(y))) -> PLUS₂(s₁(x), y)
ACK₂(s₁(x), s₁(y)) -> PLUS₂(y, ack₂(s₁(x), y))
PLUS₂(s₁(s₁(x)), y) -> PLUS₂(x, s₁(y))
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

plus₂(s₁(s₁(x)), y) -> s₁(plus₂(x, s₁(y)))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(s₁(0), y) -> s₁(y)
plus₂(0, y) -> y
ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, plus₂(y, ack₂(s₁(x), y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

PLUS₂(x, s₁(s₁(y))) -> PLUS₂(s₁(x), y)
PLUS₂(s₁(s₁(x)), y) -> PLUS₂(x, s₁(y))

The TRS R consists of the following rules:

plus₂(s₁(s₁(x)), y) -> s₁(plus₂(x, s₁(y)))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(s₁(0), y) -> s₁(y)
plus₂(0, y) -> y
ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, plus₂(y, ack₂(s₁(x), y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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, plus₂(y, ack₂(s₁(x), y)))
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

plus₂(s₁(s₁(x)), y) -> s₁(plus₂(x, s₁(y)))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(s₁(0), y) -> s₁(y)
plus₂(0, y) -> y
ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, plus₂(y, ack₂(s₁(x), y)))

Q is empty.
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, plus₂(y, 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)
plus₂(x1, x2) = plus₂(x1, x2)
ack₂(x1, x2) = ack₁(x2)
0 = 0

Lexicographic Path Order [19].
Precedence:

plus₂ > s₁
0 > ack₁ > s₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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:

plus₂(s₁(s₁(x)), y) -> s₁(plus₂(x, s₁(y)))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(s₁(0), y) -> s₁(y)
plus₂(0, y) -> y
ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, plus₂(y, ack₂(s₁(x), y)))

Q is empty.
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:

[ACK₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

plus₂(s₁(s₁(x)), y) -> s₁(plus₂(x, s₁(y)))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(s₁(0), y) -> s₁(y)
plus₂(0, y) -> y
ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, plus₂(y, ack₂(s₁(x), y)))

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