Term Rewriting System R:
[X, Y]
minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(s(X), s(Y)) -> P(minus(X, Y))
MINUS(s(X), s(Y)) -> MINUS(X, Y)
DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))





The following dependency pair can be strictly oriented:

MINUS(s(X), s(Y)) -> MINUS(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(MINUS(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(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


Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))





The following dependency pair can be strictly oriented:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))


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

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X


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

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))





Using the Dependency Graph resulted in no new DP problems.

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