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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(x, c(x), c(y)) -> F(y, y, f(y, x, y))
F(x, c(x), c(y)) -> F(y, x, y)
F(s(x), y, z) -> F(x, s(c(y)), c(z))

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Instantiation Transformation
       →DP Problem 2
         ↳Inst


Dependency Pair:

F(s(x), y, z) -> F(x, s(c(y)), c(z))


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y


Strategy:

innermost




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

F(s(x), y, z) -> F(x, s(c(y)), c(z))
one new Dependency Pair is created:

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 3
             ↳Instantiation Transformation
       →DP Problem 2
         ↳Inst


Dependency Pair:

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y


Strategy:

innermost




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

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))
one new Dependency Pair is created:

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 4
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Inst


Dependency Pair:

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c)=  0  
  POL(s(x1))=  1 + x1  
  POL(F(x1, x2, x3))=  x1 + x2 + x3  

resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2, x3) -> F(x1, x2, x3)
s(x1) -> s(x1)
c(x1) -> c


   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳Inst


Dependency Pair:


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Instantiation Transformation


Dependency Pair:

F(x, c(x), c(y)) -> F(y, x, y)


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y


Strategy:

innermost




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

F(x, c(x), c(y)) -> F(y, x, y)
one new Dependency Pair is created:

F(c(y''), c(c(y'')), c(y'')) -> F(y'', c(y''), y'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Forward Instantiation Transformation


Dependency Pair:

F(c(y''), c(c(y'')), c(y'')) -> F(y'', c(y''), y'')


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y


Strategy:

innermost




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

F(c(y''), c(c(y'')), c(y'')) -> F(y'', c(y''), y'')
one new Dependency Pair is created:

F(c(c(y''''')), c(c(c(y'''''))), c(c(y'''''))) -> F(c(y'''''), c(c(y''''')), c(y'''''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Argument Filtering and Ordering


Dependency Pair:

F(c(c(y''''')), c(c(c(y'''''))), c(c(y'''''))) -> F(c(y'''''), c(c(y''''')), c(y'''''))


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(c(c(y''''')), c(c(c(y'''''))), c(c(y'''''))) -> F(c(y'''''), c(c(y''''')), c(y'''''))


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1))=  1 + x1  
  POL(F(x1, x2, x3))=  1 + x1 + x2 + x3  

resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2, x3) -> F(x1, x2, x3)
c(x1) -> c(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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