Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/AG01SLASHHASH3.22.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:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

Q is empty.

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

TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, s₁(z))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, plus₂(y, times₂(s₁(z), 0)))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(s₁(z), 0)
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)
TIMES₂(x, s₁(y)) -> PLUS₂(times₂(x, y), x)
TIMES₂(x, plus₂(y, s₁(z))) -> PLUS₂(y, times₂(s₁(z), 0))
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)
TIMES₂(x, plus₂(y, s₁(z))) -> PLUS₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))

The TRS R consists of the following rules:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(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:

TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, s₁(z))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, plus₂(y, times₂(s₁(z), 0)))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(s₁(z), 0)
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)
TIMES₂(x, s₁(y)) -> PLUS₂(times₂(x, y), x)
TIMES₂(x, plus₂(y, s₁(z))) -> PLUS₂(y, times₂(s₁(z), 0))
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)
TIMES₂(x, plus₂(y, s₁(z))) -> PLUS₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))

The TRS R consists of the following rules:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(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.

PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

Used argument filtering: PLUS₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

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

TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, s₁(z))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, plus₂(y, times₂(s₁(z), 0)))
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)

The TRS R consists of the following rules:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(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.

TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, s₁(z))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, plus₂(y, times₂(s₁(z), 0)))
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)

Used argument filtering: TIMES₂(x1, x2) = x2
plus₂(x1, x2) = plus₂(x1, x2)
s₁(x1) = s₁(x1)
times₂(x1, x2) = times
0 = 0
Used ordering: Precedence:

plus₂ > s₁ > times > 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(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.