Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/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)))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

HALF₁(s₁(s₁(x))) -> HALF₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
HALF₁(x1) = HALF₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > HALF₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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

                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

LOG₁(s₁(s₁(x))) -> LOG₁(s₁(half₁(x)))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LOG₁(x1) = LOG₁(x1)
s₁(x1) = s₁(x1)
half₁(x1) = half₁(x1)
0 = 0

Lexicographic Path Order [19].
Precedence:

LOG₁ > half₁
s₁ > half₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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.