Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/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))

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)))

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 [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LE₂(x1, x2) = LE₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

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

                ↳ 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)))

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
IFMINUS₃(x1, x2, x3) = IFMINUS₂(x2, x3)
false = false
s₁(x1) = s₁(x1)
MINUS₂(x1, x2) = MINUS₂(x1, x2)
le₂(x1, x2) = le₂(x1, x2)
0 = 0
true = true

Lexicographic Path Order [19].
Precedence:

false > [IFMINUS₂, MINUS₂, le_2] > s₁
[0, true]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

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 [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT₂(x1, x2) = QUOT₁(x1)
s₁(x1) = s₁(x1)
minus₂(x1, x2) = x1
le₂(x1, x2) = x2
0 = 0
true = true
ifMinus₃(x1, x2, x3) = x2
false = false

Lexicographic Path Order [19].
Precedence:

true > 0
false > [QUOT₁, s_1] > 0

The following usable rules [14] were oriented:

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))

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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:

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.