Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.

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_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 [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:

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


GE_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( GE_ACTIVE2(x1, x2) ) = 3x2 + 3


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:

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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MINUS_ACTIVE2(s1(x), s1(y)) -> MINUS_ACTIVE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS_ACTIVE2(x1, x2) ) = 3x2 + 3


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:

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
            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(if3(x, y, z)) -> MARK1(x)
MARK1(if3(x, y, z)) -> IF_ACTIVE3(mark1(x), y, z)
The remaining pairs can at least be oriented weakly.

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)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( minus2(x1, x2) ) = max{0, -2}


POL( mark1(x1) ) = max{0, 2x1 - 3}


POL( if3(x1, ..., x3) ) = x1 + x2 + 2x3 + 1


POL( IF_ACTIVE3(x1, ..., x3) ) = 2x2 + 2x3


POL( MARK1(x1) ) = 2x1


POL( DIV_ACTIVE2(x1, x2) ) = max{0, -3}


POL( 0 ) = 0


POL( div2(x1, x2) ) = 2x1


POL( minus_active2(x1, x2) ) = max{0, 2x1 + 3x2 - 3}


POL( ge2(x1, x2) ) = 3x1 + 3x2 + 1


POL( true ) = max{0, -3}


POL( ge_active2(x1, x2) ) = 0


POL( false ) = max{0, -3}


POL( s1(x1) ) = 2x1


POL( div_active2(x1, x2) ) = max{0, 3x2 - 3}


POL( if_active3(x1, ..., x3) ) = x1 + 3x2 + 3x3 + 3



The following usable rules [14] were oriented: none



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

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

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)
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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(s1(x)) -> MARK1(x)
MARK1(div2(x, y)) -> MARK1(x)
The remaining pairs can at least be oriented weakly.

IF_ACTIVE3(true, x, y) -> MARK1(x)
MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
IF_ACTIVE3(false, x, y) -> MARK1(y)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( minus2(x1, x2) ) = 0


POL( mark1(x1) ) = 2x1


POL( if3(x1, ..., x3) ) = 2x2 + 3x3


POL( IF_ACTIVE3(x1, ..., x3) ) = max{0, 2x1 + x2 + x3 - 2}


POL( MARK1(x1) ) = x1 + 2


POL( DIV_ACTIVE2(x1, x2) ) = x1 + 3


POL( 0 ) = max{0, -3}


POL( div2(x1, x2) ) = 2x1 + 1


POL( minus_active2(x1, x2) ) = max{0, -2}


POL( ge2(x1, x2) ) = 1


POL( true ) = 2


POL( ge_active2(x1, x2) ) = 2


POL( false ) = 2


POL( s1(x1) ) = x1 + 1


POL( div_active2(x1, x2) ) = 2x1 + 2


POL( if_active3(x1, ..., x3) ) = 2x2 + 3x3



The following usable rules [14] were oriented:

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



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

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

IF_ACTIVE3(true, x, y) -> MARK1(x)
MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
IF_ACTIVE3(false, x, y) -> MARK1(y)
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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
The remaining pairs can at least be oriented weakly.

IF_ACTIVE3(true, x, y) -> MARK1(x)
IF_ACTIVE3(false, x, y) -> MARK1(y)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( minus2(x1, x2) ) = max{0, -2}


POL( mark1(x1) ) = max{0, -3}


POL( if3(x1, ..., x3) ) = 3x1 + 3x2 + 3x3 + 3


POL( IF_ACTIVE3(x1, ..., x3) ) = 2x2 + 3x3


POL( MARK1(x1) ) = max{0, 2x1 - 2}


POL( DIV_ACTIVE2(x1, x2) ) = max{0, -3}


POL( 0 ) = max{0, -3}


POL( div2(x1, x2) ) = 3


POL( minus_active2(x1, x2) ) = 0


POL( ge2(x1, x2) ) = 3x1 + 3x2 + 2


POL( true ) = 1


POL( ge_active2(x1, x2) ) = max{0, -3}


POL( false ) = 2


POL( s1(x1) ) = max{0, -3}


POL( div_active2(x1, x2) ) = max{0, -3}


POL( if_active3(x1, ..., x3) ) = 3x2



The following usable rules [14] were oriented: none



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

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

IF_ACTIVE3(true, x, y) -> MARK1(x)
IF_ACTIVE3(false, x, y) -> MARK1(y)
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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 3 less nodes.