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

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
         ↳Size-Change Principle
       →DP Problem 2
         ↳MRR


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





We number the DPs as follows:
  1. G(s(x)) -> G(x)
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳SCP
       →DP Problem 2
         ↳Modular Removal of Rules


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





We have the following set of usable rules:

g(s(x)) -> s(g(x))
g(0) -> 0
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(0)=  0  
  POL(g(x1))=  x1  
  POL(s(x1))=  x1  
  POL(F(x1, x2, x3))=  1 + x1 + x2 + x3  

We have the following set D of usable symbols: {0, g, s, F}
No Dependency Pairs can be deleted.
1 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳SCP
       →DP Problem 2
         ↳MRR
           →DP Problem 3
             ↳Non Termination


Dependency Pair:

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


Rules:


g(s(x)) -> s(g(x))
g(0) -> 0





Found an infinite P-chain over R:
P =

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

R =

g(s(x)) -> s(g(x))
g(0) -> 0

s = F(g(s(0)), g(s(0)), g(s(0)))
evaluates to t =F(g(s(0)), g(s(0)), g(s(0)))

Thus, s starts an infinite chain.

Non-Termination of R could be shown.
Duration:
0:00 minutes