Term Rewriting System R:
[x, y, z]
f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(f(a, b), x) -> F(a, f(a, x))
F(f(a, b), x) -> F(a, x)
F(f(b, a), x) -> F(b, f(b, x))
F(f(b, a), x) -> F(b, x)
F(x, f(y, z)) -> F(f(x, y), z)
F(x, f(y, z)) -> F(x, y)

Furthermore, R contains one SCC. 


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


Dependency Pairs:

F(f(b, a), x) -> F(b, x)
F(f(b, a), x) -> F(b, f(b, x))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, x)
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), x) -> F(a, f(a, x))


Rules:


f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




We have the following set of usable rules:

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

We have the following set D of usable symbols: {b, a, F, f}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

F(f(b, a), x) -> F(b, x)
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, 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
             ↳Instantiation Transformation


Dependency Pairs:

F(f(b, a), x) -> F(b, f(b, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), x) -> F(a, f(a, x))


Rules:


f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




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

F(x, f(y, z)) -> F(f(x, y), z)
two new Dependency Pairs are created:

F(a, f(y', z')) -> F(f(a, y'), z')
F(b, f(y', z')) -> F(f(b, y'), z')

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 3
                 ↳Semantic Labelling


Dependency Pairs:

F(b, f(y', z')) -> F(f(b, y'), z')
F(f(b, a), x) -> F(b, f(b, x))


Rules:


f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




Using Semantic Labelling, the DP problem could be transformed. The following model was found:
F(x0, x1)=  0
f(x0, x1)=  x0
b=  1
a=  0

From the dependency graph we obtain 0 (labeled) SCCs which each result in correspondingDP problem.


   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 4
                 ↳Semantic Labelling


Dependency Pairs:

F(a, f(y', z')) -> F(f(a, y'), z')
F(f(a, b), x) -> F(a, f(a, x))


Rules:


f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)


Strategy:

innermost




Using Semantic Labelling, the DP problem could be transformed. The following model was found:
F(x0, x1)=  0
a=  0
f(x0, x1)=  x0
b=  1

From the dependency graph we obtain 0 (labeled) SCCs which each result in correspondingDP problem.

Innermost Termination of R successfully shown.
Duration:
0:16 minutes