Termination w.r.t. Q proof of TRS/Rubiodivision.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:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
minus₂(0, Y) -> 0
minus₂(s₁(X), Y) -> ifMinus₃(le₂(s₁(X), Y), s₁(X), Y)
ifMinus₃(true, s₁(X), Y) -> 0
ifMinus₃(false, s₁(X), Y) -> s₁(minus₂(X, Y))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
minus₂(0, Y) -> 0
minus₂(s₁(X), Y) -> ifMinus₃(le₂(s₁(X), Y), s₁(X), Y)
ifMinus₃(true, s₁(X), Y) -> 0
ifMinus₃(false, s₁(X), Y) -> s₁(minus₂(X, Y))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, 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)) -> QUOT₂(minus₂(X, Y), s₁(Y))
IFMINUS₃(false, s₁(X), Y) -> MINUS₂(X, Y)
MINUS₂(s₁(X), Y) -> LE₂(s₁(X), Y)
MINUS₂(s₁(X), Y) -> IFMINUS₃(le₂(s₁(X), Y), s₁(X), Y)
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
QUOT₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
minus₂(0, Y) -> 0
minus₂(s₁(X), Y) -> ifMinus₃(le₂(s₁(X), Y), s₁(X), Y)
ifMinus₃(true, s₁(X), Y) -> 0
ifMinus₃(false, s₁(X), Y) -> s₁(minus₂(X, Y))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, 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)) -> QUOT₂(minus₂(X, Y), s₁(Y))
IFMINUS₃(false, s₁(X), Y) -> MINUS₂(X, Y)
MINUS₂(s₁(X), Y) -> LE₂(s₁(X), Y)
MINUS₂(s₁(X), Y) -> IFMINUS₃(le₂(s₁(X), Y), s₁(X), Y)
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
QUOT₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
minus₂(0, Y) -> 0
minus₂(s₁(X), Y) -> ifMinus₃(le₂(s₁(X), Y), s₁(X), Y)
ifMinus₃(true, s₁(X), Y) -> 0
ifMinus₃(false, s₁(X), Y) -> s₁(minus₂(X, Y))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, 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 2 less nodes.

↳ 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:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
minus₂(0, Y) -> 0
minus₂(s₁(X), Y) -> ifMinus₃(le₂(s₁(X), Y), s₁(X), Y)
ifMinus₃(true, s₁(X), Y) -> 0
ifMinus₃(false, s₁(X), Y) -> s₁(minus₂(X, Y))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, 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·x₂
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:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
minus₂(0, Y) -> 0
minus₂(s₁(X), Y) -> ifMinus₃(le₂(s₁(X), Y), s₁(X), Y)
ifMinus₃(true, s₁(X), Y) -> 0
ifMinus₃(false, s₁(X), Y) -> s₁(minus₂(X, Y))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, 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:

IFMINUS₃(false, s₁(X), Y) -> MINUS₂(X, Y)
MINUS₂(s₁(X), Y) -> IFMINUS₃(le₂(s₁(X), Y), s₁(X), Y)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
minus₂(0, Y) -> 0
minus₂(s₁(X), Y) -> ifMinus₃(le₂(s₁(X), Y), s₁(X), Y)
ifMinus₃(true, s₁(X), Y) -> 0
ifMinus₃(false, s₁(X), Y) -> s₁(minus₂(X, Y))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, 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.

IFMINUS₃(false, s₁(X), Y) -> MINUS₂(X, Y)

The remaining pairs can at least be oriented weakly.

MINUS₂(s₁(X), Y) -> IFMINUS₃(le₂(s₁(X), Y), s₁(X), Y)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(IFMINUS₃(x₁, x₂, x₃)) = x₂
POL(MINUS₂(x₁, x₂)) = x₁
POL(false) = 0
POL(le₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 1 + x₁
POL(true) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

MINUS₂(s₁(X), Y) -> IFMINUS₃(le₂(s₁(X), Y), s₁(X), Y)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
minus₂(0, Y) -> 0
minus₂(s₁(X), Y) -> ifMinus₃(le₂(s₁(X), Y), s₁(X), Y)
ifMinus₃(true, s₁(X), Y) -> 0
ifMinus₃(false, s₁(X), Y) -> s₁(minus₂(X, Y))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, 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 0 SCCs with 1 less node.

↳ 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(X, Y), s₁(Y))

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
minus₂(0, Y) -> 0
minus₂(s₁(X), Y) -> ifMinus₃(le₂(s₁(X), Y), s₁(X), Y)
ifMinus₃(true, s₁(X), Y) -> 0
ifMinus₃(false, s₁(X), Y) -> s₁(minus₂(X, Y))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))

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