Termination w.r.t. Q proof of /tmp/tmpd6bBKY/Ex49

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:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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:

DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
MINUS(s(X), s(Y)) → MINUS(X, Y)
GEQ(s(X), s(Y)) → GEQ(X, Y)
DIV(s(X), s(Y)) → MINUS(X, Y)
DIV(s(X), s(Y)) → GEQ(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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:

DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
MINUS(s(X), s(Y)) → MINUS(X, Y)
GEQ(s(X), s(Y)) → GEQ(X, Y)
DIV(s(X), s(Y)) → MINUS(X, Y)
DIV(s(X), s(Y)) → GEQ(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

GEQ(s(X), s(Y)) → GEQ(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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.

GEQ(s(X), s(Y)) → GEQ(X, Y)

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

POL(s(x₁)) = 1/2 + (4)x_1
POL(GEQ(x₁, x₂)) = (1/2)x_1

The value of delta used in the strict ordering is 1/4.
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:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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:

MINUS(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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.

MINUS(s(X), s(Y)) → MINUS(X, Y)

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

POL(MINUS(x₁, x₂)) = x_2
POL(s(x₁)) = 1 + x_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:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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:

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

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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.

DIV(s(X), s(Y)) → DIV(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)) = 0
POL(DIV(x₁, x₂)) = (4)x_1
POL(s(x₁)) = 1/4 + (1/2)x_1
POL(0) = 0

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

minus(0, Y) → 0
minus(s(X), s(Y)) → 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:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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.