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

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)) -> *'(x, y)
*'(*(i(x), k(y, z)), x) -> K(*(*(i(x), y), x), *(*(i(x), z), x))
*'(*(i(x), k(y, z)), x) -> *'(*(i(x), y), x)
*'(*(i(x), k(y, z)), x) -> *'(i(x), y)
*'(*(i(x), k(y, z)), x) -> *'(*(i(x), z), x)
*'(*(i(x), k(y, z)), x) -> *'(i(x), 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:

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


Rules:


*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1


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 3
             ↳Polynomial Ordering
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

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


Rules:


*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1


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

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:


*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1


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 Pairs:

*'(*(i(x), k(y, z)), x) -> *'(i(x), z)
*'(*(i(x), k(y, z)), x) -> *'(*(i(x), z), x)
*'(x, *(y, z)) -> *'(x, y)
*'(*(i(x), k(y, z)), x) -> *'(i(x), y)
*'(*(i(x), k(y, z)), x) -> *'(*(i(x), y), x)
*'(x, *(y, z)) -> *'(*(x, y), z)


Rules:


*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1


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)
two new Dependency Pairs are created:

*'(x'', *(*(y'', z''), z)) -> *'(x'', *(y'', z''))
*'(*(i(x'''), k(y'', z'')), *(y0, z)) -> *'(*(i(x'''), k(y'', z'')), y0)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 5
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

*'(*(i(x), k(y, z)), x) -> *'(*(i(x), z), x)
*'(*(i(x'''), k(y'', z'')), *(y0, z)) -> *'(*(i(x'''), k(y'', z'')), y0)
*'(x'', *(*(y'', z''), z)) -> *'(x'', *(y'', z''))
*'(*(i(x), k(y, z)), x) -> *'(i(x), y)
*'(*(i(x), k(y, z)), x) -> *'(*(i(x), y), x)
*'(x, *(y, z)) -> *'(*(x, y), z)
*'(*(i(x), k(y, z)), x) -> *'(i(x), z)


Rules:


*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:01 minutes