Term Rewriting System R:
[x, y, z]
i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


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


Dependency Pair:

+'(x, +(y, z)) -> +'(x, y)


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x


Strategy:

innermost




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

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

+'(x'', +(+(y'', z''), z)) -> +'(x'', +(y'', z''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Polynomial Ordering
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

+'(x'', +(+(y'', z''), z)) -> +'(x'', +(y'', z''))


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x


Strategy:

innermost




The following dependency pair can be strictly oriented:

+'(x'', +(+(y'', z''), z)) -> +'(x'', +(y'', z''))


Additionally, the following usable rules for innermost can be oriented:

+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(i(x1))=  x1  
  POL(+(x1, x2))=  1 + x1 + x2  
  POL(+'(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Polo
             ...
               →DP Problem 4
                 ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst


Dependency Pair:


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

I(+(x, y)) -> I(x)


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x


Strategy:

innermost




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

I(+(x, y)) -> I(x)
one new Dependency Pair is created:

I(+(+(x'', y''), y)) -> I(+(x'', y''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 5
             ↳Polynomial Ordering


Dependency Pair:

I(+(+(x'', y''), y)) -> I(+(x'', y''))


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x


Strategy:

innermost




The following dependency pair can be strictly oriented:

I(+(+(x'', y''), y)) -> I(+(x'', y''))


Additionally, the following usable rules for innermost can be oriented:

+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  x1  
  POL(0)=  0  
  POL(i(x1))=  x1  
  POL(+(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 5
             ↳Polo
             ...
               →DP Problem 6
                 ↳Dependency Graph


Dependency Pair:


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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