First slide

Introduction to mathematical reasoning

Question

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

Moderate

A

a contradiction
B

a tautology
C

neither a tautology nor a contradiction
D

both a tautology and a contradiction

Solution

$\begin{array}{rcl} If (p \land ~ q)^{\land} (~ p \land q) & \equiv & (p \land ~ p)^{\land} (~ q \land q) \\ \begin{matrix} \equiv (f \land f) \\ \equiv f \end{matrix} \end{array}$
$(f \to false)$
$(By using associative laws and commutative laws)$
$∴ (p \land ~ q) \land (~ p \land q) is a contradiction.$

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