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

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

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:

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

The set Q consists of the following terms:

min₂(x0, 0)
min₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
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:

MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)
QUOT₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)
LOG₁(s₁(s₁(X))) -> QUOT₂(X, s₁(s₁(0)))
QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(min₂(X, Y), s₁(Y))
LOG₁(s₁(s₁(X))) -> LOG₁(s₁(quot₂(X, s₁(s₁(0)))))

The TRS R consists of the following rules:

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

The set Q consists of the following terms:

min₂(x0, 0)
min₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
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:

MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)
QUOT₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)
LOG₁(s₁(s₁(X))) -> QUOT₂(X, s₁(s₁(0)))
QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(min₂(X, Y), s₁(Y))
LOG₁(s₁(s₁(X))) -> LOG₁(s₁(quot₂(X, s₁(s₁(0)))))

The TRS R consists of the following rules:

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

The set Q consists of the following terms:

min₂(x0, 0)
min₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
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 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)

The TRS R consists of the following rules:

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

The set Q consists of the following terms:

min₂(x0, 0)
min₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
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.

MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MIN₂(x1, x2) = x2
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

The set Q consists of the following terms:

min₂(x0, 0)
min₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
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

              ↳ QDP

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

QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(min₂(X, Y), s₁(Y))

The TRS R consists of the following rules:

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

The set Q consists of the following terms:

min₂(x0, 0)
min₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
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.

QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(min₂(X, Y), s₁(Y))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT₂(x1, x2) = x1
s₁(x1) = s₁(x1)
min₂(x1, x2) = min₁(x1)
0 = 0

Lexicographic Path Order [19].
Precedence:

s₁ > min₁
0 > min₁

The following usable rules [14] were oriented:

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

The set Q consists of the following terms:

min₂(x0, 0)
min₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
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

              ↳ QDP

                ↳ QDPOrderProof

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

LOG₁(s₁(s₁(X))) -> LOG₁(s₁(quot₂(X, s₁(s₁(0)))))

The TRS R consists of the following rules:

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

The set Q consists of the following terms:

min₂(x0, 0)
min₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
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(quot₂(X, s₁(s₁(0)))))

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)
quot₂(x1, x2) = x1
0 = 0
min₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented:

quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(X))) -> s₁(log₁(s₁(quot₂(X, s₁(s₁(0))))))

The set Q consists of the following terms:

min₂(x0, 0)
min₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
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.