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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
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))

The set Q consists of the following terms:

ge2(x0, 0)
ge2(0, s1(x0))
ge2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(s1(x0), s1(x1))
div2(x0, x1)
ify3(false, x0, x1)
ify3(true, x0, x1)
if3(false, x0, x1)
if3(true, x0, x1)


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))

The set Q consists of the following terms:

ge2(x0, 0)
ge2(0, s1(x0))
ge2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(s1(x0), s1(x1))
div2(x0, x1)
ify3(false, x0, x1)
ify3(true, x0, x1)
if3(false, x0, x1)
if3(true, x0, x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

ge2(x0, 0)
ge2(0, s1(x0))
ge2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(s1(x0), s1(x1))
div2(x0, x1)
ify3(false, x0, x1)
ify3(true, x0, x1)
if3(false, x0, x1)
if3(true, x0, x1)

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
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

ge2(x0, 0)
ge2(0, s1(x0))
ge2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(s1(x0), s1(x1))
div2(x0, x1)
ify3(false, x0, x1)
ify3(true, x0, x1)
if3(false, x0, x1)
if3(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS2(x1, x2)  =  MINUS1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

ge2(x0, 0)
ge2(0, s1(x0))
ge2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(s1(x0), s1(x1))
div2(x0, x1)
ify3(false, x0, x1)
ify3(true, x0, x1)
if3(false, x0, x1)
if3(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

ge2(x0, 0)
ge2(0, s1(x0))
ge2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(s1(x0), s1(x1))
div2(x0, x1)
ify3(false, x0, x1)
ify3(true, x0, x1)
if3(false, x0, x1)
if3(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


GE2(s1(x), s1(y)) -> GE2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
GE2(x1, x2)  =  GE1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

ge2(x0, 0)
ge2(0, s1(x0))
ge2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(s1(x0), s1(x1))
div2(x0, x1)
ify3(false, x0, x1)
ify3(true, x0, x1)
if3(false, x0, x1)
if3(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

ge2(x0, 0)
ge2(0, s1(x0))
ge2(s1(x0), s1(x1))
minus2(x0, 0)
minus2(s1(x0), s1(x1))
div2(x0, x1)
ify3(false, x0, x1)
ify3(true, x0, x1)
if3(false, x0, x1)
if3(true, x0, x1)

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