Is $ [(A \Rightarrow C) \land (B \Rightarrow C)] \Leftrightarrow [(A \lor B) \Rightarrow C]$ true

logic

Is it true that the two statements are equivalent:

$A$ implies $C$ and B implies $C$, $(A \Rightarrow C) \wedge (B \Rightarrow C)$;

$A$ or $B$, implies $C$, $(A \lor B) \Rightarrow C$;

where $A$, $B$, and $C$ are statements?

My attempt of a truth-table suggests so (T means true, F false).
enter image description here

As with my last question, I, with no profound mathematical knowledge whatsoever, stumbled across this issue trying some issues just for fun (now from "How to prove it", Velleman 2006). One solution for an exercise gave $[(A \lor B) \Rightarrow C]$. I arrived at $[(A \Rightarrow C) \wedge (B \Rightarrow C)]$ (The problem reads: "If $x$ is divisible by either 4 or 6, then it is not prime."). Now, I would like to know, if I went wrong, which is likely, in particular, as I am having a hard time to get my head around the "implies"-operator. As I do not know how to show equivalence of statements, I thought to give a truth-table a try.

If equivalence indeed holds: Is there a law? Blindfoldly generalizing, it seems, logical connectives get flipped.

Edit:
Due to the comment by Ronald Wong (thank you), I now also tried the following;
$(A \Rightarrow C) \wedge (B \Rightarrow C)$
$ \Leftrightarrow (\neg A \lor C) \land (\neg B \lor C)$
Then, maybe we can use distributive laws to "factorize" C;
$ \Leftrightarrow (\neg A \wedge \neg B) \lor C$
Next, it might be time for one a the very few laws I could recall: DeMorgan's law;
$ \Leftrightarrow \neg (A \lor B) \lor C$
Finally, we do the same thing as in the beginning;
$ \Leftrightarrow (A \lor B) \Rightarrow C.$

Best Answer

Yes.

enter image description here

Here is a link to the software I used.

This is known as the method of analytic tableaux.

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