Term Rewriting System R:
[Y, X, X1, X2, X3]
aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AMINUS(s(X), s(Y)) -> AMINUS(X, Y)
AGEQ(s(X), s(Y)) -> AGEQ(X, Y)
ADIV(s(X), s(Y)) -> AIF(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
ADIV(s(X), s(Y)) -> AGEQ(X, Y)
AIF(true, X, Y) -> MARK(X)
AIF(false, X, Y) -> MARK(Y)
MARK(minus(X1, X2)) -> AMINUS(X1, X2)
MARK(geq(X1, X2)) -> AGEQ(X1, X2)
MARK(div(X1, X2)) -> ADIV(mark(X1), X2)
MARK(div(X1, X2)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo


Dependency Pair:

AMINUS(s(X), s(Y)) -> AMINUS(X, Y)


Rules:


aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false





The following dependency pair can be strictly oriented:

AMINUS(s(X), s(Y)) -> AMINUS(X, Y)


Additionally, the following rules can be oriented:

aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(geq(x1, x2))=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  0  
  POL(true)=  0  
  POL(mark(x1))=  x1  
  POL(a__div(x1, x2))=  x1  
  POL(if(x1, x2, x3))=  x2 + x3  
  POL(0)=  0  
  POL(a__if(x1, x2, x3))=  x2 + x3  
  POL(a__minus(x1, x2))=  0  
  POL(A__MINUS(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  
  POL(a__geq(x1, x2))=  0  
  POL(div(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo


Dependency Pair:


Rules:


aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering
       →DP Problem 3
         ↳Polo


Dependency Pair:

AGEQ(s(X), s(Y)) -> AGEQ(X, Y)


Rules:


aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false





The following dependency pair can be strictly oriented:

AGEQ(s(X), s(Y)) -> AGEQ(X, Y)


Additionally, the following rules can be oriented:

aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(A__GEQ(x1, x2))=  1 + x1  
  POL(geq(x1, x2))=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  0  
  POL(true)=  0  
  POL(mark(x1))=  x1  
  POL(a__div(x1, x2))=  x1  
  POL(if(x1, x2, x3))=  x2 + x3  
  POL(0)=  0  
  POL(a__if(x1, x2, x3))=  x2 + x3  
  POL(a__minus(x1, x2))=  0  
  POL(s(x1))=  1 + x1  
  POL(a__geq(x1, x2))=  0  
  POL(div(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo


Dependency Pair:


Rules:


aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polynomial Ordering


Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(if(X1, X2, X3)) -> MARK(X1)
AIF(false, X, Y) -> MARK(Y)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), X2, X3)
MARK(div(X1, X2)) -> MARK(X1)
MARK(div(X1, X2)) -> ADIV(mark(X1), X2)
AIF(true, X, Y) -> MARK(X)
ADIV(s(X), s(Y)) -> AIF(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)


Rules:


aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false





The following dependency pair can be strictly oriented:

MARK(div(X1, X2)) -> MARK(X1)


Additionally, the following rules can be oriented:

aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(geq(x1, x2))=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  0  
  POL(true)=  0  
  POL(mark(x1))=  x1  
  POL(a__div(x1, x2))=  1 + x1  
  POL(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(0)=  0  
  POL(a__if(x1, x2, x3))=  x1 + x2 + x3  
  POL(a__minus(x1, x2))=  0  
  POL(s(x1))=  x1  
  POL(a__geq(x1, x2))=  0  
  POL(div(x1, x2))=  1 + x1  
  POL(A__DIV(x1, x2))=  1  
  POL(A__IF(x1, x2, x3))=  x2 + x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 6
             ↳Polynomial Ordering


Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(if(X1, X2, X3)) -> MARK(X1)
AIF(false, X, Y) -> MARK(Y)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), X2, X3)
MARK(div(X1, X2)) -> ADIV(mark(X1), X2)
AIF(true, X, Y) -> MARK(X)
ADIV(s(X), s(Y)) -> AIF(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)


Rules:


aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false





The following dependency pairs can be strictly oriented:

MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), X2, X3)


Additionally, the following rules can be oriented:

aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(geq(x1, x2))=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  0  
  POL(true)=  0  
  POL(mark(x1))=  1 + x1  
  POL(a__div(x1, x2))=  1  
  POL(if(x1, x2, x3))=  1 + x1 + x2 + x3  
  POL(0)=  0  
  POL(a__if(x1, x2, x3))=  1 + x1 + x2 + x3  
  POL(a__minus(x1, x2))=  0  
  POL(s(x1))=  x1  
  POL(a__geq(x1, x2))=  0  
  POL(div(x1, x2))=  0  
  POL(A__DIV(x1, x2))=  0  
  POL(A__IF(x1, x2, x3))=  x2 + x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 6
             ↳Polo
             ...
               →DP Problem 7
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(s(X)) -> MARK(X)
AIF(false, X, Y) -> MARK(Y)
MARK(div(X1, X2)) -> ADIV(mark(X1), X2)
AIF(true, X, Y) -> MARK(X)
ADIV(s(X), s(Y)) -> AIF(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)


Rules:


aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false





The following dependency pair can be strictly oriented:

MARK(s(X)) -> MARK(X)


Additionally, the following rules can be oriented:

aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(geq(x1, x2))=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  0  
  POL(true)=  0  
  POL(mark(x1))=  x1  
  POL(a__div(x1, x2))=  x1  
  POL(if(x1, x2, x3))=  x2 + x3  
  POL(0)=  0  
  POL(a__if(x1, x2, x3))=  x2 + x3  
  POL(a__minus(x1, x2))=  0  
  POL(s(x1))=  1 + x1  
  POL(a__geq(x1, x2))=  0  
  POL(div(x1, x2))=  x1  
  POL(A__DIV(x1, x2))=  x1  
  POL(A__IF(x1, x2, x3))=  x2 + x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 6
             ↳Polo
             ...
               →DP Problem 8
                 ↳Polynomial Ordering


Dependency Pairs:

AIF(false, X, Y) -> MARK(Y)
MARK(div(X1, X2)) -> ADIV(mark(X1), X2)
AIF(true, X, Y) -> MARK(X)
ADIV(s(X), s(Y)) -> AIF(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)


Rules:


aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false





The following dependency pair can be strictly oriented:

MARK(div(X1, X2)) -> ADIV(mark(X1), X2)


Additionally, the following rules can be oriented:

aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(geq(x1, x2))=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  0  
  POL(true)=  0  
  POL(mark(x1))=  1  
  POL(a__div(x1, x2))=  1  
  POL(if(x1, x2, x3))=  0  
  POL(0)=  0  
  POL(a__if(x1, x2, x3))=  1  
  POL(a__minus(x1, x2))=  0  
  POL(s(x1))=  0  
  POL(a__geq(x1, x2))=  0  
  POL(div(x1, x2))=  1  
  POL(A__DIV(x1, x2))=  0  
  POL(A__IF(x1, x2, x3))=  x2 + x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 6
             ↳Polo
             ...
               →DP Problem 9
                 ↳Dependency Graph


Dependency Pairs:

AIF(false, X, Y) -> MARK(Y)
AIF(true, X, Y) -> MARK(X)
ADIV(s(X), s(Y)) -> AIF(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)


Rules:


aminus(0, Y) -> 0
aminus(s(X), s(Y)) -> aminus(X, Y)
aminus(X1, X2) -> minus(X1, X2)
ageq(X, 0) -> true
ageq(0, s(Y)) -> false
ageq(s(X), s(Y)) -> ageq(X, Y)
ageq(X1, X2) -> geq(X1, X2)
adiv(0, s(Y)) -> 0
adiv(s(X), s(Y)) -> aif(ageq(X, Y), s(div(minus(X, Y), s(Y))), 0)
adiv(X1, X2) -> div(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> aminus(X1, X2)
mark(geq(X1, X2)) -> ageq(X1, X2)
mark(div(X1, X2)) -> adiv(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
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