Term Rewriting System R:
[x, y, z]
:(x, x) -> e
:(x, e) -> x
:(:(x, y), z) -> :(x, :(z, i(y)))
:(e, x) -> i(x)
:(x, :(y, i(x))) -> i(y)
:(x, :(y, :(i(x), z))) -> :(i(z), y)
:(i(x), :(y, x)) -> i(y)
:(i(x), :(y, :(x, z))) -> :(i(z), y)
i(:(x, y)) -> :(y, x)
i(i(x)) -> x
i(e) -> e

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

:'(:(x, y), z) -> :'(x, :(z, i(y)))
:'(:(x, y), z) -> :'(z, i(y))
:'(:(x, y), z) -> I(y)
:'(e, x) -> I(x)
:'(x, :(y, i(x))) -> I(y)
:'(x, :(y, :(i(x), z))) -> :'(i(z), y)
:'(x, :(y, :(i(x), z))) -> I(z)
:'(i(x), :(y, x)) -> I(y)
:'(i(x), :(y, :(x, z))) -> :'(i(z), y)
:'(i(x), :(y, :(x, z))) -> I(z)
I(:(x, y)) -> :'(y, x)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Modular Removal of Rules


Dependency Pairs:

:'(i(x), :(y, :(x, z))) -> I(z)
:'(i(x), :(y, :(x, z))) -> :'(i(z), y)
:'(x, :(y, :(i(x), z))) -> I(z)
:'(e, x) -> I(x)
I(:(x, y)) -> :'(y, x)
:'(:(x, y), z) -> I(y)
:'(:(x, y), z) -> :'(z, i(y))
:'(x, :(y, :(i(x), z))) -> :'(i(z), y)
:'(:(x, y), z) -> :'(x, :(z, i(y)))


Rules:


:(x, x) -> e
:(x, e) -> x
:(:(x, y), z) -> :(x, :(z, i(y)))
:(e, x) -> i(x)
:(x, :(y, i(x))) -> i(y)
:(x, :(y, :(i(x), z))) -> :(i(z), y)
:(i(x), :(y, x)) -> i(y)
:(i(x), :(y, :(x, z))) -> :(i(z), y)
i(:(x, y)) -> :(y, x)
i(i(x)) -> x
i(e) -> e


Strategy:

innermost




We have the following set of usable rules:

:(i(x), :(y, :(x, z))) -> :(i(z), y)
:(x, x) -> e
i(:(x, y)) -> :(y, x)
i(e) -> e
:(e, x) -> i(x)
:(x, :(y, :(i(x), z))) -> :(i(z), y)
i(i(x)) -> x
:(:(x, y), z) -> :(x, :(z, i(y)))
:(x, :(y, i(x))) -> i(y)
:(x, e) -> x
:(i(x), :(y, x)) -> i(y)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(:(x1, x2))=  1 + x1 + x2  
  POL(I(x1))=  x1  
  POL(i(x1))=  x1  
  POL(e)=  0  
  POL(:'(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {:, I, i, e, :'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

:'(i(x), :(y, :(x, z))) -> I(z)
:'(i(x), :(y, :(x, z))) -> :'(i(z), y)
:'(x, :(y, :(i(x), z))) -> I(z)
I(:(x, y)) -> :'(y, x)
:'(:(x, y), z) -> I(y)
:'(:(x, y), z) -> :'(z, i(y))
:'(x, :(y, :(i(x), z))) -> :'(i(z), y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

:(i(x), :(y, :(x, z))) -> :(i(z), y)
:(x, x) -> e
:(e, x) -> i(x)
:(x, :(y, :(i(x), z))) -> :(i(z), y)
:(x, :(y, i(x))) -> i(y)
:(x, e) -> x
:(i(x), :(y, x)) -> i(y)


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳Modular Removal of Rules


Dependency Pairs:

:'(e, x) -> I(x)
:'(:(x, y), z) -> :'(x, :(z, i(y)))


Rules:


i(:(x, y)) -> :(y, x)
i(e) -> e
i(i(x)) -> x
:(:(x, y), z) -> :(x, :(z, i(y)))


Strategy:

innermost




We have the following set of usable rules:

i(:(x, y)) -> :(y, x)
i(e) -> e
i(i(x)) -> x
:(:(x, y), z) -> :(x, :(z, i(y)))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(:(x1, x2))=  x1 + x2  
  POL(I(x1))=  x1  
  POL(i(x1))=  x1  
  POL(e)=  0  
  POL(:'(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {:, I, i, e, :'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

:'(e, x) -> I(x)

No Rules can be deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 3
                 ↳Dependency Graph


Dependency Pair:

:'(:(x, y), z) -> :'(x, :(z, i(y)))


Rules:


i(:(x, y)) -> :(y, x)
i(e) -> e
i(i(x)) -> x
:(:(x, y), z) -> :(x, :(z, i(y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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