Term Rewriting System R:
[x, y, z]
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))

Innermost 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
         ↳Narrowing Transformation


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))


Strategy:

innermost




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

IF(if(x, y, z), u, v) -> IF(x, if(y, u, v), if(z, u, v))
six new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


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))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:03 minutes