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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

F(f(0, x), 1) -> F(x, x)
one new Dependency Pair is created:

F(f(0, g(x'')), 1) -> F(g(x''), g(x''))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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


Rules:


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


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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 3
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

F(f(0, x), 1) -> F(g(f(x, x)), x)
one new Dependency Pair is created:

F(f(0, g(x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(f(0, g(x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))
F(f(0, g(x'')), 1) -> F(g(x''), g(x''))


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

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


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

resulting in one new DP problem. 



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


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 6
                 ↳Polynomial Ordering


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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