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

Question

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 $C\implies B$?

marlu · Accepted Answer · 2012-06-09 11:52:04Z

12

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

answered Jun 9, 2012 at 11:52

marlu

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Dan Christensen · Accepted Answer · 2012-06-12 18:47:36Z

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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.

answered Jun 12, 2012 at 5:27

Dan Christensen

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$\begingroup$ But this should be for all $C$ $\endgroup$
– Belgi
Jun 12, 2012 at 5:43
$\begingroup$ It is for all C. $\endgroup$
– Dan Christensen
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 $B$. $\endgroup$
– Carl Mummert
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. $\endgroup$
– Dan Christensen
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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