JEE-Mains 2015 (29/30)

Logic Level 2

The negation of $\sim s \lor ( \sim r \land s)$ is equivalent to $\text{\_\_\_\_\_\_\_\_\_\_}.$

s \land r

s \land (r \land \sim s)

s \lor (r \lor \sim s)

s \land \sim r

1 solution

Jake Lai
Jun 20, 2015

We recall that De Morgan's Law states that

$\neg (p \land q) \equiv \neg p \lor \neg q \quad ; \quad \neg (p \lor q) \equiv \neg p \land \neg q$

Along with $p \equiv \neg \neg p$ and the distributive property, we thus have

$\neg (\neg s \lor (\neg r \land s)) \equiv s \land \neg (\neg r \land s) \equiv s \land (r \lor \neg s) \equiv (s \land r) \lor (s \land \neg s) \equiv (s \land r) \lor F$

where $s \land \neg s \equiv F$ is a contradiction. Since contradictions are never true, we have the statement equivalent to

$\boxed{s \land r}$

Moderator note:

Good job working through the propositional logic.

For this particular problem, to target solving in JEE, it is much faster to draw a 2-circle Venn Diagram, and the answer would be obvious.

JEE-Mains 2015 (29/30)

1 solution

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