$(p\vee q)$ $\wedge$ $(p\wedge q)' \equiv ? $
I wanna find what this equivalent to. So my teacher has solved this but If it's possible, wanna get a solution which i can understand easily.
My idea (I mean my teacher :P):
$\equiv p \vee q$
Regards!
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1First parenthesis says at least one, and the second says not both. This is called "exclusive or": $p\veebar q$. See https://en.wikipedia.org/wiki/Exclusive_or – Coolwater Sep 22 '17 at 18:01
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My idea is that it is equivalent to itself. – Sep 22 '17 at 18:02
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I didn't get what you mean – Lan Sep 22 '17 at 18:02
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lol, my teacher found that answer so i wanna get a good solution which i can understand. – Lan Sep 22 '17 at 18:03
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Your teacher is wrong. – Sep 22 '17 at 18:04
3 Answers
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$$(p \lor q) \land (p \land q)'\equiv (p \lor q)\land (p' \lor q')\tag{DeMorgan's}$$
Now, distribute:
$$\begin{align}(p \lor q)\land (p' \lor q')& \equiv \Big((p \land (p'\lor q')) \lor (q \land (p'\lor q'))\Big)\\ \\ & \equiv [\underbrace{(p \wedge p')}_{\bot = 0}\vee (p \wedge q')]\vee [(q \wedge p')\vee (\underbrace{q\wedge q')}_{\bot = 0}]\tag{Distribution again}\\ \\ & \equiv (p \land q') \lor (p' \land q)\end{align}$$
This is, if you're familiar with it, $p \operatorname{XOR} q$
amWhy
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\begin{eqnarray*} (p\vee q) \wedge (p\wedge q)' &\equiv & (p\vee q) \wedge (p'\vee q')\\ & \equiv& [p \wedge (p'\vee q')]\vee [q \wedge (p'\vee q')]\\ & \equiv & [(p \wedge p')\vee (p \wedge q')]\vee [(q \wedge p')\vee (q\wedge q')] \\ & \equiv & (p \wedge q')\vee (q \wedge p') \end{eqnarray*}
nonuser
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