Term Rewriting System R:
[x, y, z]
f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(0, 1, x) -> F(g(x), g(x), x)
F(g(x), y, z) -> F(x, y, z)
F(x, g(y), z) -> F(x, y, z)
F(x, y, g(z)) -> F(x, y, z)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

F(x, y, g(z)) -> F(x, y, z)
F(x, g(y), z) -> F(x, y, z)
F(g(x), y, z) -> F(x, y, z)
F(0, 1, x) -> F(g(x), g(x), x)


Rules:


f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(x, y, g(z)) -> F(x, y, z)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(g(x1))=  1 + x1  
  POL(1)=  0  
  POL(F(x1, x2, x3))=  x3  

resulting in one new DP problem. 



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


Dependency Pairs:

F(x, g(y), z) -> F(x, y, z)
F(g(x), y, z) -> F(x, y, z)
F(0, 1, x) -> F(g(x), g(x), x)


Rules:


f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))


Strategy:

innermost




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

F(g(x), y, z) -> F(x, y, z)
three new Dependency Pairs are created:

F(g(g(x'')), y'', z'') -> F(g(x''), y'', z'')
F(g(0), 1, z') -> F(0, 1, z')
F(g(x''), g(y''), z'') -> F(x'', g(y''), z'')

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

F(g(x''), g(y''), z'') -> F(x'', g(y''), z'')
F(g(0), 1, z') -> F(0, 1, z')
F(g(g(x'')), y'', z'') -> F(g(x''), y'', z'')
F(0, 1, x) -> F(g(x), g(x), x)
F(x, g(y), z) -> F(x, y, z)


Rules:


f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))


Strategy:

innermost




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

F(x, g(y), z) -> F(x, y, z)
five new Dependency Pairs are created:

F(x'', g(g(y'')), z'') -> F(x'', g(y''), z'')
F(0, g(1), z') -> F(0, 1, z')
F(g(g(x'''')), g(y'), z') -> F(g(g(x'''')), y', z')
F(g(0), g(1), z') -> F(g(0), 1, z')
F(g(x''''), g(g(y'''')), z') -> F(g(x''''), g(y''''), z')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

F(g(0), g(1), z') -> F(g(0), 1, z')
F(g(x''''), g(g(y'''')), z') -> F(g(x''''), g(y''''), z')
F(g(g(x'''')), g(y'), z') -> F(g(g(x'''')), y', z')
F(0, g(1), z') -> F(0, 1, z')
F(x'', g(g(y'')), z'') -> F(x'', g(y''), z'')
F(0, 1, x) -> F(g(x), g(x), x)
F(g(0), 1, z') -> F(0, 1, z')
F(g(g(x'')), y'', z'') -> F(g(x''), y'', z'')
F(g(x''), g(y''), z'') -> F(x'', g(y''), z'')


Rules:


f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))


Strategy:

innermost



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