Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/RubioSLASHdivision.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:

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

  ↳ Non-Overlap Check

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.

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:

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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
ifMinus₃(true, s₁(x0), x1)
ifMinus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))

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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
ifMinus₃(true, s₁(x0), x1)
ifMinus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))

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:

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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
ifMinus₃(true, s₁(x0), x1)
ifMinus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
ifMinus₃(true, s₁(x0), x1)
ifMinus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)

Used argument filtering: LE₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
ifMinus₃(true, s₁(x0), x1)
ifMinus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))

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

                ↳ QDPAfsSolverProof

              ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
ifMinus₃(true, s₁(x0), x1)
ifMinus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: IFMINUS₃(x1, x2, x3) = x2
s₁(x1) = s₁(x1)
MINUS₂(x1, x2) = x1
le₂(x1, x2) = le
0 = 0
false = false
true = true
Used ordering: Quasi Precedence: le > false le > true

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
ifMinus₃(true, s₁(x0), x1)
ifMinus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
ifMinus₃(true, s₁(x0), x1)
ifMinus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(minus₂(X, Y), s₁(Y))

Used argument filtering: QUOT₂(x1, x2) = x1
s₁(x1) = s₁(x1)
minus₂(x1, x2) = x1
0 = 0
ifMinus₃(x1, x2, x3) = x2
le₂(x1, x2) = le
false = false
true = true
Used ordering: Quasi Precedence: s_1 > 0 [le, false] > true > 0

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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)))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
ifMinus₃(true, s₁(x0), x1)
ifMinus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.