Term Rewriting System R:
[X, Y]
f(g(X), Y) -> f(X, f(g(X), Y))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(g(X), Y) -> F(X, f(g(X), Y))
F(g(X), Y) -> F(g(X), Y)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Negative Polynomial Order


Dependency Pairs:

F(g(X), Y) -> F(g(X), Y)
F(g(X), Y) -> F(X, f(g(X), Y))


Rule:


f(g(X), Y) -> f(X, f(g(X), Y))


Strategy:

innermost




The following Dependency Pair can be strictly oriented using the given order.

F(g(X), Y) -> F(X, f(g(X), Y))


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

f(g(X), Y) -> f(X, f(g(X), Y))


Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, x2) ) = x1

POL( g(x1) ) = x1 + 1

POL( f(x1, x2) ) = 0


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳Neg POLO
           →DP Problem 2
             ↳Usable Rules (Innermost)


Dependency Pair:

F(g(X), Y) -> F(g(X), Y)


Rule:


f(g(X), Y) -> f(X, f(g(X), Y))


Strategy:

innermost




As we are in the innermost case, we can delete all 1 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳Neg POLO
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 3
                 ↳Non Termination


Dependency Pair:

F(g(X), Y) -> F(g(X), Y)


Rule:

none


Strategy:

innermost




Found an infinite P-chain over R:
P =

F(g(X), Y) -> F(g(X), Y)

R = none

s = F(g(X'), Y')
evaluates to t =F(g(X'), Y')

Thus, s starts an infinite chain.

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