Term Rewriting System R:

   [Y, X]
   minus(0, Y) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   geq(X, 0) -> true
   geq(0, s(Y)) -> false
   geq(s(X), s(Y)) -> geq(X, Y)
   div(0, s(Y)) -> 0
   div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   if(true, X, Y) -> X
   if(false, X, Y) -> Y

Termination of R to be shown.


This program has no overlaps, so it is sufficient to show innermost termination. 


R contains the following Dependency Pairs: 


   GEQ(s(X), s(Y)) -> GEQ(X, Y)
   DIV(s(X), s(Y)) -> IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   DIV(s(X), s(Y)) -> GEQ(X, Y)
   DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
   DIV(s(X), s(Y)) -> MINUS(X, Y)
   MINUS(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains three SCCs.


SCC1:

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

Removing rules from R by ordering and analyzing Dependency Pairs, Usable Rules, and Usable Equations.

This is possible by using the following (C_E-compatible) Polynomial ordering.

Polynomial interpretation:
POL(s(x_1)) = 1 + x_1
POL(GEQ(x_1, x_2)) = 1 + x_1 + x_2

The following Dependency Pairs can be deleted:

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



 This transformation is resulting in no new subcycles.


SCC2:

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

Removing rules from R by ordering and analyzing Dependency Pairs, Usable Rules, and Usable Equations.

This is possible by using the following (C_E-compatible) Polynomial ordering.

Polynomial interpretation:
POL(s(x_1)) = 1 + x_1
POL(MINUS(x_1, x_2)) = 1 + x_1 + x_2

The following Dependency Pairs can be deleted:

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



 This transformation is resulting in no new subcycles.


SCC3:

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

On this Scc, a Narrowing SCC transformation can be performed.
 
As a result of transforming the rule 

   DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
two new Dependency Pairs
are created:


   DIV(s(0), s(Y'')) -> DIV(0, s(Y''))
   DIV(s(s(X'')), s(s(Y''))) -> DIV(minus(X'', Y''), s(s(Y'')))

The transformation is resulting in no subcycles.




Termination of R successfully shown.


Duration: 0.644 seconds.