First slide

Introduction to mathematical reasoning

Question

The negation of $(~ p \land q) \lor (p \land ~ q)$ is

Easy

A

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

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

$(p \land ~ q) \land (~ p \lor q)$
D

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

Solution

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

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