First slide

Introduction to mathematical reasoning

Question

Negation of $p \to q$ is

Easy

A

$p \land (~ q)$
B

$~ p \lor q$
C

$~ q \to\sim p$
D

$p \lor (~ q)$

Solution

$~ (p \to q) \equiv\sim (~ p \lor q) \equiv\sim (~ p) \land (~ q)$
[De Morgan's Laws]
$∴ ~ (p \to q) \equiv p \land (~ q)$

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