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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


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





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

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

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


Rules:


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





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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

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


Rules:


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





The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

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


Rules:


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





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

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

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


Rules:


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





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

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

F(c(y'''''), c(c(y''0))) -> F(c(y''0), c(y''0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳Inst
             ...
               →DP Problem 7
                 ↳Polynomial Ordering


Dependency Pair:

F(c(y'''''), c(c(y''0))) -> F(c(y''0), c(y''0))


Rules:


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





The following dependency pair can be strictly oriented:

F(c(y'''''), c(c(y''0))) -> F(c(y''0), c(y''0))


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

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

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.

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