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_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(x, y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(x, y)

Q is empty.

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

DIV_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
IF_ACTIVE3(true, x, y) -> MARK1(x)
GE_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
MARK1(if3(x, y, z)) -> MARK1(x)
MARK1(s1(x)) -> MARK1(x)
MARK1(ge2(x, y)) -> GE_ACTIVE2(x, y)
MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
MARK1(div2(x, y)) -> MARK1(x)
MARK1(minus2(x, y)) -> MINUS_ACTIVE2(x, y)
MINUS_ACTIVE2(s1(x), s1(y)) -> MINUS_ACTIVE2(x, y)
IF_ACTIVE3(false, x, y) -> MARK1(y)
MARK1(if3(x, y, z)) -> IF_ACTIVE3(mark1(x), y, z)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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:

DIV_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
IF_ACTIVE3(true, x, y) -> MARK1(x)
GE_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
MARK1(if3(x, y, z)) -> MARK1(x)
MARK1(s1(x)) -> MARK1(x)
MARK1(ge2(x, y)) -> GE_ACTIVE2(x, y)
MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
MARK1(div2(x, y)) -> MARK1(x)
MARK1(minus2(x, y)) -> MINUS_ACTIVE2(x, y)
MINUS_ACTIVE2(s1(x), s1(y)) -> MINUS_ACTIVE2(x, y)
IF_ACTIVE3(false, x, y) -> MARK1(y)
MARK1(if3(x, y, z)) -> IF_ACTIVE3(mark1(x), y, z)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

GE_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

GE_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
Used argument filtering: GE_ACTIVE2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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
            ↳ QDPAfsSolverProof
          ↳ QDP

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

MINUS_ACTIVE2(s1(x), s1(y)) -> MINUS_ACTIVE2(x, y)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS_ACTIVE2(s1(x), s1(y)) -> MINUS_ACTIVE2(x, y)
Used argument filtering: MINUS_ACTIVE2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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

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

MARK1(if3(x, y, z)) -> MARK1(x)
MARK1(s1(x)) -> MARK1(x)
IF_ACTIVE3(true, x, y) -> MARK1(x)
MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
MARK1(div2(x, y)) -> MARK1(x)
IF_ACTIVE3(false, x, y) -> MARK1(y)
MARK1(if3(x, y, z)) -> IF_ACTIVE3(mark1(x), y, z)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.