Prove $A \implies C$

Question

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.

Answer 1

No, you need to apply Distribution before Simplification. Simplification is not an equivalence, so cannot safely be applied to only part of a statement.

$$\begin{align} &(A\lor B)\to C &&\text{Premise} \\ \iff & \lnot(A \lor B)\lor C&& \text{Conditional Exchange}\\\iff & (\lnot A\land\lnot B)\lor C&&\text{de Morgan's}\\\iff &(\lnot A\lor C)\land(\lnot B\lor C)&&\text{Distribution}\\\implies &\lnot A\lor C&&\text{Simplification}\\ \iff & A\to C&&\text{Conditional Exchange} \end{align}$$

Answer 2

DeMorgan's yes, since it is an equivalence. Simplification, not necessarily since it takes something to a weaker statement. (It is justified here since weakening a disjunct leads to a weaker statement, but it won't be true in general that weakening a subformula leads to a weaker statement.)

Instead, use distribution to get $(\lnot A\lor C)\land (\lnot B\lor C),$ and then use simplification to get $\lnot A\lor C.$