Termination w.r.t. Q proof of /tmp/tmp1q96F4/division.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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))
IFMINUS(false, s(X), Y) → MINUS(X, Y)
MINUS(s(X), Y) → LE(s(X), Y)
MINUS(s(X), Y) → IFMINUS(le(s(X), Y), s(X), Y)
LE(s(X), s(Y)) → LE(X, Y)
QUOT(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(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:

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))
IFMINUS(false, s(X), Y) → MINUS(X, Y)
MINUS(s(X), Y) → LE(s(X), Y)
MINUS(s(X), Y) → IFMINUS(le(s(X), Y), s(X), Y)
LE(s(X), s(Y)) → LE(X, Y)
QUOT(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

LE(s(X), s(Y)) → LE(X, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

LE(s(X), s(Y)) → LE(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x₁)) = 4 + (2)x_1
POL(LE(x₁, x₂)) = (3)x_2

The value of delta used in the strict ordering is 12.
The following usable rules [17] 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(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:

IFMINUS(false, s(X), Y) → MINUS(X, Y)
MINUS(s(X), Y) → IFMINUS(le(s(X), Y), s(X), Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

IFMINUS(false, s(X), Y) → MINUS(X, Y)
MINUS(s(X), Y) → IFMINUS(le(s(X), Y), s(X), Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(IFMINUS(x₁, x₂, x₃)) = 2 + (2)x_2 + (4)x_3
POL(le(x₁, x₂)) = 3 + x_1 + (3)x_2
POL(true) = 2
POL(false) = 2
POL(MINUS(x₁, x₂)) = 3 + (2)x_1 + (4)x_2
POL(s(x₁)) = 3 + x_1
POL(0) = 1

The value of delta used in the strict ordering is 1.
The following usable rules [17] 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(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

            ↳ QDPOrderProof

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

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(minus(x₁, x₂)) = 7/4 + (2)x_1
POL(ifMinus(x₁, x₂, x₃)) = 1/2 + (2)x_2
POL(QUOT(x₁, x₂)) = (2)x_1
POL(le(x₁, x₂)) = 0
POL(true) = 2
POL(false) = 0
POL(s(x₁)) = 15/4 + (2)x_1
POL(0) = 3/2

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(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.