Termination w.r.t. Q proof of TRS/AG01#3.8a.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:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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:

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
MINUS₂(x, s₁(y)) -> PRED₁(minus₂(x, y))
LOG₁(s₁(s₁(x))) -> QUOT₂(x, s₁(s₁(0)))
LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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:

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
MINUS₂(x, s₁(y)) -> PRED₁(minus₂(x, y))
LOG₁(s₁(s₁(x))) -> QUOT₂(x, s₁(s₁(0)))
LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

MINUS₂(x, s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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.

MINUS₂(x, s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS₂(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( s₁(x₁) ) = 3x₁ + 1

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:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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(minus₂(x, y), s₁(y))

The TRS R consists of the following rules:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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(minus₂(x, y), s₁(y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = 2x₁ + 3

POL( 0 ) = 0

POL( minus₂(x₁, x₂) ) = x₁

POL( pred₁(x₁) ) = max{0, x₁ - 3}

POL( QUOT₂(x₁, x₂) ) = max{0, 3x₁ + 3x₂ - 3}

The following usable rules [14] were oriented:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))

↳ 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:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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 Order [17,21] with Interpretation:

POL( s₁(x₁) ) = 3x₁ + 3

POL( 0 ) = max{0, -2}

POL( minus₂(x₁, x₂) ) = x₁

POL( pred₁(x₁) ) = max{0, x₁ - 3}

POL( LOG₁(x₁) ) = 2x₁ + 2

POL( quot₂(x₁, x₂) ) = 2x₁ + 1

The following usable rules [14] were oriented:

pred₁(s₁(x)) -> x
quot₂(0, s₁(y)) -> 0
minus₂(x, 0) -> x
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))

↳ 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:

pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(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.