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
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

MINUS_ACTIVE(s(x), s(y)) → MINUS_ACTIVE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ 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]:

POL(0) = 0   
POL(DIV_ACTIVE(x1, x2)) = 0   
POL(IF_ACTIVE(x1, x2, x3)) = x2 + x3   
POL(MARK(x1)) = x1   
POL(div(x1, x2)) = x1   
POL(div_active(x1, x2)) = 0   
POL(false) = 0   
POL(ge(x1, x2)) = 0   
POL(ge_active(x1, x2)) = 0   
POL(if(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(if_active(x1, x2, x3)) = 0   
POL(mark(x1)) = 0   
POL(minus(x1, x2)) = 0   
POL(minus_active(x1, x2)) = 0   
POL(s(x1)) = x1   
POL(true) = 0   

The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ Instantiation

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.
By instantiating [15] the rule IF_ACTIVE(false, x, y) → MARK(y) we obtained the following new rules:

IF_ACTIVE(false, s(div(minus(y_3, y_4), s(y_5))), 0) → MARK(0)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ Instantiation
QDP
                    ↳ DependencyGraphProof

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

IF_ACTIVE(false, s(div(minus(y_3, y_4), s(y_5))), 0) → MARK(0)
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)

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 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ Instantiation
                  ↳ QDP
                    ↳ DependencyGraphProof
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(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)

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)
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)
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(DIV_ACTIVE(x1, x2)) = 1   
POL(IF_ACTIVE(x1, x2, x3)) = x2   
POL(MARK(x1)) = x1   
POL(div(x1, x2)) = 1 + x1   
POL(div_active(x1, x2)) = 0   
POL(false) = 0   
POL(ge(x1, x2)) = 0   
POL(ge_active(x1, x2)) = 0   
POL(if(x1, x2, x3)) = 0   
POL(if_active(x1, x2, x3)) = 0   
POL(mark(x1)) = 0   
POL(minus(x1, x2)) = 0   
POL(minus_active(x1, x2)) = 0   
POL(s(x1)) = x1   
POL(true) = 0   

The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ Instantiation
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ QDPOrderProof
QDP
                            ↳ QDPOrderProof

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

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)

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(s(x)) → 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)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(DIV_ACTIVE(x1, x2)) = x1   
POL(IF_ACTIVE(x1, x2, x3)) = x2   
POL(MARK(x1)) = x1   
POL(div(x1, x2)) = x1   
POL(div_active(x1, x2)) = x1   
POL(false) = 0   
POL(ge(x1, x2)) = 0   
POL(ge_active(x1, x2)) = 0   
POL(if(x1, x2, x3)) = x1 + x2 + x3   
POL(if_active(x1, x2, x3)) = x1 + x2 + x3   
POL(mark(x1)) = x1   
POL(minus(x1, x2)) = 0   
POL(minus_active(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [17] were oriented:

mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
minus_active(0, y) → 0
ge_active(0, s(y)) → false
mark(minus(x, y)) → minus_active(x, y)
ge_active(x, 0) → true
mark(s(x)) → s(mark(x))
div_active(0, s(y)) → 0
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)
mark(ge(x, y)) → ge_active(x, y)
minus_active(x, y) → minus(x, y)
div_active(x, y) → div(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
                ↳ Instantiation
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
QDP
                                ↳ Instantiation

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

MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
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)

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.
By instantiating [15] the rule IF_ACTIVE(true, x, y) → MARK(x) we obtained the following new rules:

IF_ACTIVE(true, s(div(minus(y_3, y_4), s(y_5))), 0) → MARK(s(div(minus(y_3, y_4), s(y_5))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ Instantiation
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ Instantiation
QDP
                                    ↳ DependencyGraphProof

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

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)
IF_ACTIVE(true, s(div(minus(y_3, y_4), s(y_5))), 0) → MARK(s(div(minus(y_3, y_4), s(y_5))))

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 3 less nodes.