Term Rewriting System R:
[x, y]
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
LT(s(x), s(y)) -> LT(x, y)
DIV(s(x), s(y)) -> IF(lt(x, y), 0, s(div(-(x, y), s(y))))
DIV(s(x), s(y)) -> LT(x, y)
DIV(s(x), s(y)) -> DIV(-(x, y), s(y))
DIV(s(x), s(y)) -> -'(x, y)

Furthermore, R contains three SCCs. 


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


Dependency Pair:

-'(s(x), s(y)) -> -'(x, y)


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))





The following dependency pair can be strictly oriented:

-'(s(x), s(y)) -> -'(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(-'(x1, x2))=  x1  
  POL(s(x1))=  1 + 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:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))





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:

LT(s(x), s(y)) -> LT(x, y)


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))





The following dependency pair can be strictly oriented:

LT(s(x), s(y)) -> LT(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(LT(x1, x2))=  x1  
  POL(s(x1))=  1 + 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:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))





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 Pair:

DIV(s(x), s(y)) -> DIV(-(x, y), s(y))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))





The following dependency pair can be strictly oriented:

DIV(s(x), s(y)) -> DIV(-(x, y), s(y))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(DIV(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  
  POL(-(x1, x2))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes