Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/AG01SLASHHASH3.7.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:

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(half₁(x))))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(half₁(x))))

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:

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(half₁(x))))

The set Q consists of the following terms:

half₁(0)
half₁(s₁(s₁(x0)))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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

LOG₁(s₁(s₁(x))) -> HALF₁(x)
LOG₁(s₁(s₁(x))) -> LOG₁(s₁(half₁(x)))
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(half₁(x))))

The set Q consists of the following terms:

half₁(0)
half₁(s₁(s₁(x0)))
log₁(s₁(0))
log₁(s₁(s₁(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:

LOG₁(s₁(s₁(x))) -> HALF₁(x)
LOG₁(s₁(s₁(x))) -> LOG₁(s₁(half₁(x)))
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(half₁(x))))

The set Q consists of the following terms:

half₁(0)
half₁(s₁(s₁(x0)))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(half₁(x))))

The set Q consists of the following terms:

half₁(0)
half₁(s₁(s₁(x0)))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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.

HALF₁(s₁(s₁(x))) -> HALF₁(x)

Used argument filtering: HALF₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(half₁(x))))

The set Q consists of the following terms:

half₁(0)
half₁(s₁(s₁(x0)))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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

LOG₁(s₁(s₁(x))) -> LOG₁(s₁(half₁(x)))

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(half₁(x))))

The set Q consists of the following terms:

half₁(0)
half₁(s₁(s₁(x0)))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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.

LOG₁(s₁(s₁(x))) -> LOG₁(s₁(half₁(x)))

Used argument filtering: LOG₁(x1) = x1
s₁(x1) = s₁(x1)
half₁(x1) = x1
0 = 0
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(half₁(x))))

The set Q consists of the following terms:

half₁(0)
half₁(s₁(s₁(x0)))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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