Termination w.r.t. Q proof of /tmp/tmpiv1-HZ/JFP

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_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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:

GE_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)
MINUS_ACTIVE(s(x), s(y)) → MINUS_ACTIVE(x, y)
MARK(ge(x, y)) → GE_ACTIVE(x, y)
MARK(s(x)) → MARK(x)
IF_ACTIVE(true, x, y) → MARK(x)
MARK(minus(x, y)) → MINUS_ACTIVE(x, y)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
MARK(if(x, y, z)) → IF_ACTIVE(mark(x), y, z)
DIV_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)
MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
MARK(div(x, y)) → MARK(x)
MARK(if(x, y, z)) → MARK(x)
IF_ACTIVE(false, x, y) → MARK(y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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:

GE_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)
MINUS_ACTIVE(s(x), s(y)) → MINUS_ACTIVE(x, y)
MARK(ge(x, y)) → GE_ACTIVE(x, y)
MARK(s(x)) → MARK(x)
IF_ACTIVE(true, x, y) → MARK(x)
MARK(minus(x, y)) → MINUS_ACTIVE(x, y)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
MARK(if(x, y, z)) → IF_ACTIVE(mark(x), y, z)
DIV_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)
MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
MARK(div(x, y)) → MARK(x)
MARK(if(x, y, z)) → MARK(x)
IF_ACTIVE(false, x, y) → MARK(y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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:

GE_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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.

GE_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)

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

POL(GE_ACTIVE(x₁, x₂)) = (15/4)x_2
POL(s(x₁)) = 1/2 + (13/4)x_1

The value of delta used in the strict ordering is 15/8.
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_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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_ACTIVE(s(x), s(y)) → MINUS_ACTIVE(x, y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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_ACTIVE(s(x), s(y)) → MINUS_ACTIVE(x, y)

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

POL(MINUS_ACTIVE(x₁, x₂)) = (15/4)x_2
POL(s(x₁)) = 1/2 + (13/4)x_1

The value of delta used in the strict ordering is 15/8.
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_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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:

MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
MARK(div(x, y)) → MARK(x)
MARK(if(x, y, z)) → MARK(x)
MARK(s(x)) → MARK(x)
IF_ACTIVE(true, x, y) → MARK(x)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
MARK(if(x, y, z)) → IF_ACTIVE(mark(x), y, z)
IF_ACTIVE(false, x, y) → MARK(y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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.

MARK(if(x, y, z)) → MARK(x)
MARK(if(x, y, z)) → IF_ACTIVE(mark(x), y, z)

The remaining pairs can at least be oriented weakly.

MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
MARK(div(x, y)) → MARK(x)
MARK(s(x)) → MARK(x)
IF_ACTIVE(true, x, y) → MARK(x)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
IF_ACTIVE(false, x, y) → MARK(y)

Used ordering: Polynomial interpretation [25,35]:

POL(IF_ACTIVE(x₁, x₂, x₃)) = (2)x_2 + x_3
POL(if_active(x₁, x₂, x₃)) = 11/4 + (1/4)x_2 + (15/4)x_3
POL(ge_active(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = 0
POL(true) = 0
POL(mark(x₁)) = 0
POL(DIV_ACTIVE(x₁, x₂)) = 0
POL(0) = 0
POL(minus_active(x₁, x₂)) = 15/4 + (4)x_2
POL(MARK(x₁)) = x_1
POL(if(x₁, x₂, x₃)) = 1/4 + x_1 + (2)x_2 + x_3
POL(div_active(x₁, x₂)) = 1 + (1/4)x_2
POL(div(x₁, x₂)) = (4)x_1
POL(false) = 0
POL(s(x₁)) = (4)x_1
POL(ge(x₁, x₂)) = (11/4)x_2

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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
MARK(div(x, y)) → MARK(x)
IF_ACTIVE(true, x, y) → MARK(x)
MARK(s(x)) → MARK(x)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
IF_ACTIVE(false, x, y) → MARK(y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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.

MARK(div(x, y)) → MARK(x)

The remaining pairs can at least be oriented weakly.

MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
IF_ACTIVE(true, x, y) → MARK(x)
MARK(s(x)) → MARK(x)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
IF_ACTIVE(false, x, y) → MARK(y)

Used ordering: Polynomial interpretation [25,35]:

POL(IF_ACTIVE(x₁, x₂, x₃)) = (1/4)x_2 + (1/2)x_3
POL(if_active(x₁, x₂, x₃)) = 4 + (4)x_1 + x_2 + (4)x_3
POL(ge_active(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = 0
POL(true) = 0
POL(mark(x₁)) = 0
POL(DIV_ACTIVE(x₁, x₂)) = 1/4
POL(0) = 0
POL(minus_active(x₁, x₂)) = 4 + (3/4)x_2
POL(if(x₁, x₂, x₃)) = 3/2 + (4)x_1 + (7/4)x_2 + (15/4)x_3
POL(MARK(x₁)) = (1/4)x_1
POL(div_active(x₁, x₂)) = 4 + (5/4)x_1
POL(div(x₁, x₂)) = 1 + (4)x_1
POL(false) = 0
POL(s(x₁)) = x_1
POL(ge(x₁, x₂)) = 4 + (3)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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
MARK(s(x)) → MARK(x)
IF_ACTIVE(true, x, y) → MARK(x)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
IF_ACTIVE(false, x, y) → MARK(y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, 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.

MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
MARK(s(x)) → MARK(x)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)

The remaining pairs can at least be oriented weakly.

IF_ACTIVE(true, x, y) → MARK(x)
IF_ACTIVE(false, x, y) → MARK(y)

Used ordering: Polynomial interpretation [25,35]:

POL(IF_ACTIVE(x₁, x₂, x₃)) = 3/4 + (1/4)x_2 + (1/4)x_3
POL(if_active(x₁, x₂, x₃)) = 1/2 + x_2 + (4)x_3
POL(ge_active(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = 0
POL(true) = 0
POL(mark(x₁)) = x_1
POL(DIV_ACTIVE(x₁, x₂)) = 3/4 + (3/4)x_1 + (1/4)x_2
POL(0) = 0
POL(minus_active(x₁, x₂)) = 0
POL(if(x₁, x₂, x₃)) = 1/2 + x_2 + (4)x_3
POL(MARK(x₁)) = 3/4 + (1/4)x_1
POL(div_active(x₁, x₂)) = 1/4 + (4)x_1 + x_2
POL(div(x₁, x₂)) = 1/4 + (4)x_1 + x_2
POL(false) = 0
POL(s(x₁)) = 1/4 + x_1
POL(ge(x₁, x₂)) = 0

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

div_active(x, y) → div(x, y)
minus_active(0, y) → 0
mark(0) → 0
mark(s(x)) → s(mark(x))
minus_active(s(x), s(y)) → minus_active(x, y)
mark(minus(x, y)) → minus_active(x, y)
ge_active(x, 0) → true
mark(ge(x, y)) → ge_active(x, y)
ge_active(0, s(y)) → false
if_active(true, x, y) → mark(x)
if_active(false, x, y) → mark(y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
mark(if(x, y, z)) → if_active(mark(x), y, z)
ge_active(s(x), s(y)) → ge_active(x, y)
div_active(0, s(y)) → 0
minus_active(x, y) → minus(x, y)
if_active(x, y, z) → if(x, y, z)
ge_active(x, y) → ge(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

IF_ACTIVE(true, x, y) → MARK(x)
IF_ACTIVE(false, x, y) → MARK(y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.