First slide

Introduction to mathematical reasoning

Question

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

Easy

A

a tautology
B

a contradiction
C

both a tautology and a contradiction
D

neither a tautology nor a contradiction

Solution

$(p \land ~ q) \land (~ p \land q) = (p \land ~ p) \land (~ q \land q) = f \land f = f$
(by using associative laws and commutative laws)
$∴ (p \land ~ q) \land (~ p \land q)$ is a contradiction.

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