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:

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

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:

DIVISION2(x, y) -> DIV3(x, y, 0)
DIV3(x, y, z) -> LT2(x, y)
LT2(s1(x), s1(y)) -> LT2(x, y)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
INC1(s1(x)) -> INC1(x)
DIV3(x, y, z) -> INC1(z)
IF4(false, x, s1(y), z) -> DIV3(minus2(x, s1(y)), s1(y), z)
DIV3(x, y, z) -> IF4(lt2(x, y), x, y, inc1(z))
IF4(false, x, s1(y), z) -> MINUS2(x, s1(y))

The TRS R consists of the following rules:

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

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:

DIVISION2(x, y) -> DIV3(x, y, 0)
DIV3(x, y, z) -> LT2(x, y)
LT2(s1(x), s1(y)) -> LT2(x, y)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
INC1(s1(x)) -> INC1(x)
DIV3(x, y, z) -> INC1(z)
IF4(false, x, s1(y), z) -> DIV3(minus2(x, s1(y)), s1(y), z)
DIV3(x, y, z) -> IF4(lt2(x, y), x, y, inc1(z))
IF4(false, x, s1(y), z) -> MINUS2(x, s1(y))

The TRS R consists of the following rules:

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 4 less nodes.

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

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

INC1(s1(x)) -> INC1(x)

The TRS R consists of the following rules:

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

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.


INC1(s1(x)) -> INC1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( INC1(x1) ) = x1


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

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
          ↳ QDP

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

LT2(s1(x), s1(y)) -> LT2(x, y)

The TRS R consists of the following rules:

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

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.


LT2(s1(x), s1(y)) -> LT2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( LT2(x1, x2) ) = x1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

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
            ↳ QDPOrderProof
          ↳ 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:

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

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) ) = x1


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

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
          ↳ QDP

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

DIV3(x, y, z) -> IF4(lt2(x, y), x, y, inc1(z))
IF4(false, x, s1(y), z) -> DIV3(minus2(x, s1(y)), s1(y), z)

The TRS R consists of the following rules:

division2(x, y) -> div3(x, y, 0)
div3(x, y, z) -> if4(lt2(x, y), x, y, inc1(z))
if4(true, x, y, z) -> z
if4(false, x, s1(y), z) -> div3(minus2(x, s1(y)), s1(y), z)
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))

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