Term Rewriting System R:

   [x, y, z]
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x))) -> s(half(x))
   half(double(x)) -> x
   -(x, 0) -> x
   -(s(x), s(y)) -> -(x, y)
   if(0, y, z) -> y
   if(s(x), y, z) -> z

Termination of R to be shown.


Removing the following rules from R which fullfill a polynomial ordering: 

   double(0) -> 0
   half(double(x)) -> x
   -(x, 0) -> x
   if(0, y, z) -> y
   if(s(x), y, z) -> z

where the Polynomial interpretation:
POL(s(x_1)) = x_1
POL(-(x_1, x_2)) = 1 + x_1 + x_2
POL(double(x_1)) = 1 + x_1
POL(half(x_1)) = x_1
POL(0) = 0
POL(if(x_1, x_2, x_3)) = 1 + x_1 + x_2 + x_3
was used. 



Not all Rules of R can be deleted, so we still have to regard a part of R.


Removing the following rules from R which fullfill a polynomial ordering: 

   half(0) -> 0
   half(s(0)) -> 0

where the Polynomial interpretation:
POL(s(x_1)) = x_1
POL(-(x_1, x_2)) = 2 + x_1 + x_2
POL(double(x_1)) = x_1
POL(half(x_1)) = 2*x_1
POL(0) = 1
was used. 



Not all Rules of R can be deleted, so we still have to regard a part of R.


Removing the following rules from R which fullfill a polynomial ordering: 

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

where the Polynomial interpretation:
POL(s(x_1)) = 1 + x_1
POL(-(x_1, x_2)) = 1 + x_1 + x_2
POL(double(x_1)) = 2*x_1
POL(half(x_1)) = x_1
was used. 



Not all Rules of R can be deleted, so we still have to regard a part of R.


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


R contains the following Dependency Pairs: 


   DOUBLE(s(x)) -> DOUBLE(x)

Furthermore, R contains one SCC.


SCC1:

DOUBLE(s(x)) -> DOUBLE(x)

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(DOUBLE(x_1)) = 1 + x_1

The following Dependency Pairs can be deleted:

   DOUBLE(s(x)) -> DOUBLE(x)



 This transformation is resulting in no new subcycles.




Termination of R successfully shown.


Duration: 0.505 seconds.