Term Rewriting System R:
[x, y, z, u, v]
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
         ↳Forward Instantiation 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)


Rule:


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


Strategy:

innermost




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

IF(if(x, y, z), u, v) -> IF(y, u, v)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Forward Instantiation Transformation


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 3
                 ↳Argument Filtering and Ordering


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


The following usable rule for innermost w.r.t. to the AFS can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
IF(x1, x2, x3) -> IF(x1, x2, x3)
if(x1, x2, x3) -> if(x1, x2, x3)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pair:


Rule:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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