This question is something that I've tried to solve on my own, but with no such luck. The only thing I've managed to do using laws of logic is using contrapositive, then DeMorgan's Laws, and other laws that lead me to $(B \wedge A) \wedge (\neg C \vee \neg A)$. I feel like I've done something wrong in my calculations and I need someone to confirm it and show me what I did wrong.
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Did you try using truth tables? – Victor Palea Oct 23 '20 at 18:52
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I did and it's been proven true, but I want to use the laws of logic to prove it instead of the truth table. – Daniel Jimenez Oct 23 '20 at 19:59
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Using the known fact: $$ A \implies B \equiv \neg A \vee B\space ,$$
$$ \begin{aligned}A \implies (B \vee C) &\equiv \neg A \vee B \vee C \\ (A \implies B) \vee (A \implies C) &\equiv \neg A \vee B \vee \neg A \vee C \\&\equiv \neg A \vee B \vee C\end{aligned}$$
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I'm actually confused about the middle part. I believe that you utilized the Distributive Laws, but doesn't the conjunction next to A have to be the opposite of the conjunction with B and C in order for the law to be used? – Daniel Jimenez Oct 23 '20 at 19:56
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Sorry in advance if I understood your question wrong, the longer version of the middle part is: $( \neg A \vee B) \vee ( \neg A \vee C)$ which is $\neg A \vee B \vee \neg A \vee C \equiv \neg A \vee B \vee C$ because $\neg A$ appears twice, and the second appearance won't affect the truth value of this propositional. – CSch of x Oct 23 '20 at 21:05
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1I see. I've managed to figure it out on my own. Apparently, the laws like Idempotent and Null Laws can work with negatives as well. Thanks. Your comment gave me the clue that I need. – Daniel Jimenez Oct 23 '20 at 21:26