I am new to logic and math exchange. I am taking a class that requires a deduction of these two logical premises and a conclusion. How do I use the laws of logic to deduce this conclusion?
$( \neg C \wedge S) \vee (\neg C \wedge D)$
$\neg ( S \vee A )$
$\therefore ( \neg C \wedge \neg S) \wedge \neg A$
On
By Natural Deduction
Take the premises $(\neg C\wedge S)\vee(\neg C\wedge D)\,,\neg(S\vee A)$.
From the first premise conclude $\neg C$ by disjunctive elimination (proof by cases).
From the second premise conclude $\neg S$ by contradiction elimination (proof by contradiction).
From the second premise conclude $\neg A$ likewise.
Therefore demonstrating both premises entail $(\neg C\wedge \neg S)\wedge \neg A$ by two applications of conjunctive introduction.
$$\neg ( S \vee A ) \implies \neg S \wedge \neg A$$ $$( \neg C \wedge S) \vee (\neg C \wedge D) \implies \neg C \wedge ( S \vee D ) \implies \neg C$$ Therefore $$\neg C \wedge ( \neg S \wedge \neg A ) \implies ( \neg C \wedge \neg S) \wedge \neg A$$