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 $C\implies B$?


2 Answers 2


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


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

  • $\begingroup$ But this should be for all $C$ $\endgroup$
    – Belgi
    Jun 12, 2012 at 5:43
  • $\begingroup$ It is for all C. $\endgroup$ Jun 12, 2012 at 5:52
  • $\begingroup$ This is how I interpreted the question as well. If the formula quoted here is true for all $C$ then indeed $A$ implies $B$. $\endgroup$ Jun 12, 2012 at 12:15
  • $\begingroup$ 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. $\endgroup$ Jun 12, 2012 at 13:25
  • 1
    $\begingroup$ More explicitly, there is a huge difference between $\forall C, ( P(C) \implies Q )$ and $(\forall C, P(C)) \implies Q$. $\endgroup$
    – Erick Wong
    Jun 12, 2012 at 19:56

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