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my question is in the title:

to show $A\implies B$ is it enough to show for any $C$ such that $C\implies A$ we have $C\implies B$?

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Yes but that doesn't make it easier since you could choose $C = A$.

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Do you mean $((C\implies A) \implies (C\implies B)) \implies (A\implies B)$?

From the truth table, this is false when A is true, B is false and C is false. Therefore, this formula is not true in general.

A nice truth table generator: http://mathdl.maa.org/images/upload_library/47/mcclung/index.html

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    I could be wrong, but I don't think that was what the OP was talking about.2012-06-19