Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/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.

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 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

            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: ACK₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)

Used argument filtering: ACK₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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.