Term Rewriting System R:
[x, y, z]
.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(x))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(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) -> .'(y, z)
one new Dependency Pair is created:

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


Rules:


.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))


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

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:


.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(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(y)


Rules:


.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(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(y)
one new Dependency Pair is created:

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


Rules:


.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))


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


.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(x))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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