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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


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


Dependency Pair:

G(s(x)) -> G(x)


Rules:


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


Strategy:

innermost




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

G(s(x)) -> G(x)
one new Dependency Pair is created:

G(s(s(x''))) -> G(s(x''))

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

G(s(s(x''))) -> G(s(x''))


Rules:


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


Strategy:

innermost




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

G(s(s(x''))) -> G(s(x''))
one new Dependency Pair is created:

G(s(s(s(x'''')))) -> G(s(s(x'''')))

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
         ↳Rw


Dependency Pair:

G(s(s(s(x'''')))) -> G(s(s(x'''')))


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

G(s(s(s(x'''')))) -> G(s(s(x'''')))


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(s(x1))=  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
         ↳Rw


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

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

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

The transformation is resulting in no new DP problems.


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