Term Rewriting System R:
[x, y, z, u, v]
if(true, x, y) -> x
if(false, x, y) -> y
if(x, y, y) -> y
if(if(x, y, z), u, v) -> if(x, if(y, u, v), if(z, u, v))
if(x, if(x, y, z), z) -> if(x, y, z)
if(x, y, if(x, y, z)) -> if(x, y, z)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

IF(if(x, y, z), u, v) -> IF(x, if(y, u, v), if(z, u, v))
IF(if(x, y, z), u, v) -> IF(y, u, v)
IF(if(x, y, z), u, v) -> IF(z, u, v)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

IF(if(x, y, z), u, v) -> IF(z, u, v)
IF(if(x, y, z), u, v) -> IF(y, u, v)
IF(if(x, y, z), u, v) -> IF(x, if(y, u, v), if(z, u, v))


Rules:


if(true, x, y) -> x
if(false, x, y) -> y
if(x, y, y) -> y
if(if(x, y, z), u, v) -> if(x, if(y, u, v), if(z, u, v))
if(x, if(x, y, z), z) -> if(x, y, z)
if(x, y, if(x, y, z)) -> if(x, y, z)





The following dependency pairs can be strictly oriented:

IF(if(x, y, z), u, v) -> IF(z, u, v)
IF(if(x, y, z), u, v) -> IF(y, u, v)
IF(if(x, y, z), u, v) -> IF(x, if(y, u, v), if(z, u, v))


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(if(x1, x2, x3))=  1 + x1 + x2 + x3  
  POL(false)=  0  
  POL(true)=  0  
  POL(IF(x1, x2, x3))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


if(true, x, y) -> x
if(false, x, y) -> y
if(x, y, y) -> y
if(if(x, y, z), u, v) -> if(x, if(y, u, v), if(z, u, v))
if(x, if(x, y, z), z) -> if(x, y, z)
if(x, y, if(x, y, z)) -> if(x, y, z)





Using the Dependency Graph resulted in no new DP problems.

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