Termination w.r.t. Q proof of TRS/Rubiologarquot.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

  ↳ 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))))))

Q is empty.

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))))))

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:

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))))))

Q is empty.
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

  ↳ 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))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MIN₂(x₁, x₂)) = 2·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ 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))))))

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

            ↳ 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))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(QUOT₂(x₁, x₂)) = 2·x₁
POL(min₂(x₁, x₂)) = 2·x₁
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented:

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

↳ 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))))))

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

          ↳ 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))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(LOG₁(x₁)) = 2·x₁
POL(min₂(x₁, x₂)) = x₁
POL(quot₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented:

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

↳ 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))))))

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.