# Prove $A \implies C$

Given, $$(A \lor B) \implies C$$, prove $$A \implies C$$

My Proof:

1 By Conditional Exchange,

$$\neg(A \lor B) \lor C$$

2 By DeMorgan's Law,

$$(\neg A \land \neg B) \lor C$$

3 By Simplification,

$$\neg A \lor C$$

4 By Conditional Exchange,

$$A \implies C$$

My question pertains to steps 2 and 3. I used Demorgan's and Simplification on a subformula of a premise -- can I do that? Usually, I would separate the subformulas, but I don't think I could do so in this case.

Thanks.

• Your steps look valid – J. W. Tanner Jul 26 '19 at 4:13
• Alternative Proof: By disjunction, $A \implies A \vee B$. Since $A \vee B \implies C$, and since $A \implies A \vee B$, then $A \implies C$ by hypothetical syllogism. – JavaMan Jul 26 '19 at 4:24
• My logic is rusty, so take this with a grain of salt: but I do not see the problem. You are asked to prove "If A, then C." so I do not see why you can't start by assuming $A$ is true. In math, to prove a statement of the form $P \implies Q$, it is perfectly valid to assume $P$ is true, since if $P$ is false, then $P \implies Q$ vacuously. If this is a rigorous (philosophical) logic course, then maybe I'm missing some details in logical formalism, so you could/should ask your teacher (or others here). – JavaMan Jul 26 '19 at 4:31
• This isn't circular reasoning. Circular reasoning would be assuming that "$A \implies C$ is true to prove that $A \implies C$ is true. Here, you are really proving that $A \implies C$ using cases: whether $A$ is true or not. – JavaMan Jul 26 '19 at 4:35

\begin{align} &(A\lor B)\to C &&\text{Premise} \\ \iff & \lnot(A \lor B)\lor C&& \text{Conditional Exchange}\\\iff & (\lnot A\land\lnot B)\lor C&&\text{de Morgan's}\\\iff &(\lnot A\lor C)\land(\lnot B\lor C)&&\text{Distribution}\\\implies &\lnot A\lor C&&\text{Simplification}\\ \iff & A\to C&&\text{Conditional Exchange} \end{align}
• The obvious case where substituting a weakening does not work is when the phrase is in an antecedent. $~~(A\land B)\to B$ does not imply $A\to B$. – Graham Kemp Jul 26 '19 at 4:34
Instead, use distribution to get $$(\lnot A\lor C)\land (\lnot B\lor C),$$ and then use simplification to get $$\lnot A\lor C.$$