# Question on social choice functions

We showed in class that every strongly, exactly consistent s.c.f is strongly firm (I don't know if this is the right translation - we defined strong firmness as the equivalence of $*,\alpha,\beta$ effectiveness, which simply means that $\beta$ effectiveness $\Rightarrow$ * effectiveness).

I was wondering - is the converse statement true? That is: is every strongly firm s.c.f also strongly and exactly consistent? I thought about this for a bit and it seems the answer should be negative, however I haven't been able to come up with a suitable example.

Thanks!

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Copeland's method and Kramer's method are both good counter examples. –  shay Feb 10 '11 at 23:16