# To show $A\implies B$, is that sufficient to show for all $C$ s.t. $C\implies A$ then $C\implies B$

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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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
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