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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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2 Answers 2

up vote 12 down vote accepted

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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But this should be for all $C$ –  Belgi Jun 12 '12 at 5:43
    
It is for all C. –  Dan Christensen Jun 12 '12 at 5:52
    
This is how I interpreted the question as well. If the formula quoted here is true for all $C$ then indeed $A$ implies $B$. –  Carl Mummert Jun 12 '12 at 12:15
    
But this formula is not a tautology. In the case I mention, it is false. As such, we cannot say that it is true in general. –  Dan Christensen Jun 12 '12 at 13:25
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More explicitly, there is a huge difference between $\forall C, ( P(C) \implies Q )$ and $(\forall C, P(C)) \implies Q$. –  Erick Wong Jun 12 '12 at 19:56

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