Term Rewriting System R:
[]
f(0) -> cons(0, f(s(0)))
f(s(0)) -> f(p(s(0)))
p(s(0)) -> 0

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(0) -> F(s(0))
F(s(0)) -> F(p(s(0)))
F(s(0)) -> P(s(0))

Furthermore, R contains one SCC. 


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


Dependency Pairs:

F(s(0)) -> F(p(s(0)))
F(0) -> F(s(0))


Rules:


f(0) -> cons(0, f(s(0)))
f(s(0)) -> f(p(s(0)))
p(s(0)) -> 0


Strategy:

innermost




As we are in the innermost case, we can delete all 2 non-usable-rules.


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


Dependency Pairs:

F(s(0)) -> F(p(s(0)))
F(0) -> F(s(0))


Rule:


p(s(0)) -> 0


Strategy:

innermost




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

F(s(0)) -> F(p(s(0)))
one new Dependency Pair is created:

F(s(0)) -> F(0)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 3
                 ↳Usable Rules (Innermost)


Dependency Pairs:

F(s(0)) -> F(0)
F(0) -> F(s(0))


Rule:


p(s(0)) -> 0


Strategy:

innermost




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


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 4
                 ↳Non Termination


Dependency Pairs:

F(s(0)) -> F(0)
F(0) -> F(s(0))


Rule:

none


Strategy:

innermost




Found an infinite P-chain over R:
P =

F(s(0)) -> F(0)
F(0) -> F(s(0))

R = none

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

Thus, s starts an infinite chain.

Innermost Non-Termination of R could be shown.
Duration:
0:02 minutes