Term Rewriting System R:
[X, Y]
*(X, +(Y, 1)) -> +(*(X, +(Y, *(1, 0))), X)
*(X, 1) -> X
*(X, 0) -> X
*(X, 0) -> 0

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))
*'(X, +(Y, 1)) -> *'(1, 0)

Furthermore, R contains one SCC. 


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


Dependency Pair:

*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))


Rules:


*(X, +(Y, 1)) -> +(*(X, +(Y, *(1, 0))), X)
*(X, 1) -> X
*(X, 0) -> X
*(X, 0) -> 0





We have the following set of usable rules:

*(X, 0) -> X
*(X, 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(*'(x1, x2))=  1 + x1 + x2  
  POL(1)=  0  
  POL(*(x1, x2))=  x1 + x2  
  POL(+(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {0, *', 1, *, +}
No Dependency Pairs can be deleted.
2 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



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


Dependency Pair:

*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))


Rules:


*(X, 0) -> X
*(X, 0) -> 0





Found an infinite P-chain over R:
P =

*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))

R =

*(X, 0) -> X
*(X, 0) -> 0

s = *'(X', +(Y', *(1, 0)))
evaluates to t =*'(X', +(Y', *(1, 0)))

Thus, s starts an infinite chain.

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