Termination w.r.t. Q proof of TRS/nontermin/AG01#4.30a.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(s₁(x), s₁(y)), s₁(y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(s₁(x), s₁(y)), s₁(y)))

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)) -> MINUS₂(s₁(x), s₁(y))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(s₁(x), s₁(y)), s₁(y))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(s₁(x), s₁(y)), s₁(y)))

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)) -> MINUS₂(s₁(x), s₁(y))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(s₁(x), s₁(y)), s₁(y))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(s₁(x), s₁(y)), s₁(y)))

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(s₁(x), s₁(y)), s₁(y)))

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.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

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

POL(LE₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + 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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(s₁(x), s₁(y)), s₁(y)))

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:

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(s₁(x), s₁(y)), s₁(y)))

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(s₁(x), s₁(y)) -> MINUS₂(x, y)

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

POL(MINUS₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ 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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(s₁(x), s₁(y)), s₁(y)))

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

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

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(s₁(x), s₁(y)), s₁(y))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(s₁(x), s₁(y)), s₁(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.