Question 1101290
~P v ~Q translates to "not P or not Q"
so for example, if we have these two statements
P = pirates are real
Q = quiz is tomorrow
then ~P v ~Q would translate to "pirates are not real OR the quiz is not tomorrow"


Now consider the expression ~(P v Q)
That would be "It is not the case that pirates are real or the quiz is tomorrow"
In other words if person A says "pirates are real or the quiz is tomorrow", then person B can counter with the complete opposite and say "pirates are not real AND the quiz is not tomorrow". The "or" flips to "and". We negate everything. Even the connector between the two statements. 


This is De Morgan's Law in action. We can use truth tables to see that ~P v ~Q and ~(P v Q) are not equivalent. Compare the columns highlighted in yellow. The truth values don't match up.

<img src = "https://i.imgur.com/Ynt4SC8.png">

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Here is how to solve the proof


<table border=1><tr><th>Number</th><th>Statement</th><th>Lines Used</th><th>Reason</th></tr><tr><td>1</td><td>~P</td><td></td><td></td></tr><tr><td>2</td><td>~(P v Q) &gt; ~R</td><td></td><td></td></tr><tr><td>3</td><td>~Q</td><td></td><td></td></tr><tr><td>:.</td><td>~R</td><td></td><td></td></tr><tr><td>4</td><td>~P &amp; ~Q</td><td>1,3</td><td>Conjunction</td></tr><tr><td>5</td><td>~(P v Q)</td><td>4</td><td>De Morgan's Law</td></tr><tr><td>6</td><td>~R</td><td>2,5</td><td>Modus Ponens</td></tr></table>