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 $C\implies B$?

score 12 · Accepted Answer · answered Jun 09 '12 at 11:52

12

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

answered Jun 09 '12 at 11:52

marlu

Dan Christensen · Answer 2 · 2012-06-12T18:47:36.440

3

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.

edited Jun 12 '12 at 18:47

answered Jun 12 '12 at 05:27

Dan Christensen

But this should be for all $C$ – Belgi Jun 12 '12 at 05:43
It is for all C. – Dan Christensen Jun 12 '12 at 05:52
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
...and that, boys and girls, is the difference between "If some ravens are black then all ravens are black." and "If all ravens are black then all ravens are black." :-) – Asaf Karagila Jun 12 '12 at 18:51
1

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
I could be wrong, but I don't think that was what the OP was talking about. – Dan Christensen Jun 19 '12 at 20:04

2 Answers2