First slide

Introduction to mathematical reasoning

Question

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

Easy

A

$a t a u t o \log y$
B

$a c o n t r a d i c t i o n$
C

$B o t h 1 a n d 2$
D

$n e i t h e r 1 n o r 2$

Solution

$(p \land ~ q) \land (~ p \land q)$
$p \land (~ q \land q) \land ~ p$
$(p \land ~ p) \land (~ q \land q) ≅ f \land f \equiv f (contraciction)$

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