Term Rewriting System R:
[x, y, z]
purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

PURGE(.(x, y)) -> PURGE(remove(x, y))
PURGE(.(x, y)) -> REMOVE(x, y)
REMOVE(x, .(y, z)) -> REMOVE(x, z)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

REMOVE(x, .(y, z)) -> REMOVE(x, z)


Rules:


purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))


Strategy:

innermost




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

REMOVE(x, .(y, z)) -> REMOVE(x, z)
one new Dependency Pair is created:

REMOVE(x'', .(y, .(y'', z''))) -> REMOVE(x'', .(y'', z''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Nar


Dependency Pair:

REMOVE(x'', .(y, .(y'', z''))) -> REMOVE(x'', .(y'', z''))


Rules:


purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))


Strategy:

innermost




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

REMOVE(x'', .(y, .(y'', z''))) -> REMOVE(x'', .(y'', z''))
one new Dependency Pair is created:

REMOVE(x'''', .(y, .(y''0, .(y'''', z'''')))) -> REMOVE(x'''', .(y''0, .(y'''', z'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Nar


Dependency Pair:

REMOVE(x'''', .(y, .(y''0, .(y'''', z'''')))) -> REMOVE(x'''', .(y''0, .(y'''', z'''')))


Rules:


purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

REMOVE(x'''', .(y, .(y''0, .(y'''', z'''')))) -> REMOVE(x'''', .(y''0, .(y'''', z'''')))


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(REMOVE(x1, x2))=  1 + x1 + x2  
  POL(.(x1, x2))=  1 + x1 + x2  

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


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳Nar


Dependency Pair:


Rules:


purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Narrowing Transformation


Dependency Pair:

PURGE(.(x, y)) -> PURGE(remove(x, y))


Rules:


purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))


Strategy:

innermost




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

PURGE(.(x, y)) -> PURGE(remove(x, y))
two new Dependency Pairs are created:

PURGE(.(x'', nil)) -> PURGE(nil)
PURGE(.(x'', .(y'', z'))) -> PURGE(if(=(x'', y''), remove(x'', z'), .(y'', remove(x'', z'))))

The transformation is resulting in no new DP problems.


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