Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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
      ↳ EdgeDeletionProof

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

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

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 deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
GE_ACTIVE(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
MINUS_ACTIVE(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP

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

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

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.