# Proving or disproving expression with implies operators

I'm having a hard time reducing expressions involving "implies" operators. I did some reading about the actual meaning of the "implies" operator and browse for other Q&A on this website; however, I don't know how to interpret expressions. For example, I have:

$$(C1) A \implies B$$ $$(C2) C \implies B$$

And I as a definition given in the context I know that: $$C \implies A$$

Now, I'm being asked to prove or disprove: $$a) C1 \implies C2$$ and $$b) C2 \implies C1$$

Here is where my problem/question, for a) is it valid to do the following?

$$(A \implies B) \implies (C \implies B)$$

and then (somehow):

$$A \implies C$$

Based on my definition given a above, I'm thinking I could "disprove" a) and "prove" b) but I'm not sure if logic has any foundation.

Any help would be appreciated

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Are you allowed to use truth tables, or do you actually have to derive "$C_1 \implies C_2$"? It seems like the only way you could disprove such a statement is to find a case where $C_1$ is true but $C_2$ is false. –  you Apr 8 '12 at 23:13
I think I'm allowed to; however, I don't see a straight forward way of doing for my particular scenario. I'll keep it in mind, thanks! –  Ulises Apr 9 '12 at 18:32

You know that $C\implies A$. If you also knew $C1$, you’d have $$C\implies A\quad\text{ and }\quad A\implies B\;,$$ from which you may certainly infer $C\implies B$, which is $C2$. Thus, given that $C\implies A$, you must have that $C1\implies C2$. We’ve just proved (a).

As for (b), does knowing that $C\implies A$ and $C\implies B$ guarantee that $A\implies B$? What if $A,B$, and $C$ are:

$\qquad C:\quad n$ is a multiple of $6$,

$\qquad A:\quad n$ is a multiple of $2$, and

$\qquad B:\quad n$ is a multiple of $3$?

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Does this mean that the "implies" operator is transitive? in other words if we have have: $$C \implies A and A \implies B$$ Can we infer the following? $$C \implies B$$ –  Ulises Apr 9 '12 at 14:19
yes, implication is transitive. –  Ronald Apr 9 '12 at 16:39
Thank you guys, I guess this means $$(A \implies B) \implies (C \implies B)$$ cannot be reduced to: $$A \implies C$$ –  Ulises Apr 9 '12 at 18:29