First slide

Introduction to mathematical reasoning

Question

The negation of the compound statement $(p \lor q) \land r$ is

Easy

A

$(~ p \lor ~ q) \land ~ r$
B

$(~ p \land ~ q) \lor ~ r$
C

$~ (p \lor q) \Rightarrow r$
D

$p \land q$

Solution

Let $S : (p \lor q) \land r$
$\begin{array}{ll} \Rightarrow & ~ S :\sim (p \lor q) \land r) \\ \Rightarrow & ~ S :\sim (p \lor q) \lor ~ r (De-Morgan's Laws) \\ ∴ & ~ S : (~ p \land ~ q) \lor ~ r \end{array}$

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