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:

ge2(x, 0) -> true
ge2(0, s1(x)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
div2(x, y) -> ify3(ge2(y, s1(0)), x, y)
ify3(false, x, y) -> divByZeroError
ify3(true, x, y) -> if3(ge2(x, y), x, y)
if3(false, x, y) -> 0
if3(true, x, y) -> s1(div2(minus2(x, y), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

ge2(x, 0) -> true
ge2(0, s1(x)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
div2(x, y) -> ify3(ge2(y, s1(0)), x, y)
ify3(false, x, y) -> divByZeroError
ify3(true, x, y) -> if3(ge2(x, y), x, y)
if3(false, x, y) -> 0
if3(true, x, y) -> s1(div2(minus2(x, y), 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:

IF3(true, x, y) -> DIV2(minus2(x, y), y)
IF3(true, x, y) -> MINUS2(x, y)
IFY3(true, x, y) -> GE2(x, y)
IFY3(true, x, y) -> IF3(ge2(x, y), x, y)
DIV2(x, y) -> IFY3(ge2(y, s1(0)), x, y)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
DIV2(x, y) -> GE2(y, s1(0))
GE2(s1(x), s1(y)) -> GE2(x, y)

The TRS R consists of the following rules:

ge2(x, 0) -> true
ge2(0, s1(x)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
div2(x, y) -> ify3(ge2(y, s1(0)), x, y)
ify3(false, x, y) -> divByZeroError
ify3(true, x, y) -> if3(ge2(x, y), x, y)
if3(false, x, y) -> 0
if3(true, x, y) -> s1(div2(minus2(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:

IF3(true, x, y) -> DIV2(minus2(x, y), y)
IF3(true, x, y) -> MINUS2(x, y)
IFY3(true, x, y) -> GE2(x, y)
IFY3(true, x, y) -> IF3(ge2(x, y), x, y)
DIV2(x, y) -> IFY3(ge2(y, s1(0)), x, y)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
DIV2(x, y) -> GE2(y, s1(0))
GE2(s1(x), s1(y)) -> GE2(x, y)

The TRS R consists of the following rules:

ge2(x, 0) -> true
ge2(0, s1(x)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
div2(x, y) -> ify3(ge2(y, s1(0)), x, y)
ify3(false, x, y) -> divByZeroError
ify3(true, x, y) -> if3(ge2(x, y), x, y)
if3(false, x, y) -> 0
if3(true, x, y) -> s1(div2(minus2(x, y), 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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

ge2(x, 0) -> true
ge2(0, s1(x)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
div2(x, y) -> ify3(ge2(y, s1(0)), x, y)
ify3(false, x, y) -> divByZeroError
ify3(true, x, y) -> if3(ge2(x, y), x, y)
if3(false, x, y) -> 0
if3(true, x, y) -> s1(div2(minus2(x, y), 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.


MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS2(x1, x2) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] 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:

ge2(x, 0) -> true
ge2(0, s1(x)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
div2(x, y) -> ify3(ge2(y, s1(0)), x, y)
ify3(false, x, y) -> divByZeroError
ify3(true, x, y) -> if3(ge2(x, y), x, y)
if3(false, x, y) -> 0
if3(true, x, y) -> s1(div2(minus2(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:

GE2(s1(x), s1(y)) -> GE2(x, y)

The TRS R consists of the following rules:

ge2(x, 0) -> true
ge2(0, s1(x)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
div2(x, y) -> ify3(ge2(y, s1(0)), x, y)
ify3(false, x, y) -> divByZeroError
ify3(true, x, y) -> if3(ge2(x, y), x, y)
if3(false, x, y) -> 0
if3(true, x, y) -> s1(div2(minus2(x, y), 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.


GE2(s1(x), s1(y)) -> GE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( GE2(x1, x2) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] 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:

ge2(x, 0) -> true
ge2(0, s1(x)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
div2(x, y) -> ify3(ge2(y, s1(0)), x, y)
ify3(false, x, y) -> divByZeroError
ify3(true, x, y) -> if3(ge2(x, y), x, y)
if3(false, x, y) -> 0
if3(true, x, y) -> s1(div2(minus2(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

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

IF3(true, x, y) -> DIV2(minus2(x, y), y)
IFY3(true, x, y) -> IF3(ge2(x, y), x, y)
DIV2(x, y) -> IFY3(ge2(y, s1(0)), x, y)

The TRS R consists of the following rules:

ge2(x, 0) -> true
ge2(0, s1(x)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
div2(x, y) -> ify3(ge2(y, s1(0)), x, y)
ify3(false, x, y) -> divByZeroError
ify3(true, x, y) -> if3(ge2(x, y), x, y)
if3(false, x, y) -> 0
if3(true, x, y) -> s1(div2(minus2(x, y), y))

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