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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(x), s(y)) -> F(x, y)
G(0, x) -> G(f(x, x), x)
G(0, x) -> F(x, x)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

G(0, x) -> G(f(x, x), x)


Rules:


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


Strategy:

innermost




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

G(0, x) -> G(f(x, x), x)
two new Dependency Pairs are created:

G(0, 0) -> G(s(0), 0)
G(0, s(x'')) -> G(s(f(x'', x'')), s(x''))

The transformation is resulting in no new DP problems.


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