Question 260642
# 1


Let 

r: The apartment is hot. 
q: The air conditioner is working. 
p: The temperature is 90.



By convention, the tilde symbol ~ is to denote the opposite of a given statement. So ~p denotes "not p" and it is the opposite of p. So if r: The apartment is hot, then ~r: the apartment is NOT hot. See the difference? 


So to translate "the temperature is not 90", simply start with p and negate it to get ~p. To add on "the air conditioner is working", just combine ~p with q along with the symbol *[Tex \LARGE \wedge] like so:


*[Tex \LARGE \sim p \wedge q]


Remember that *[Tex \LARGE \wedge] means "and"



Finally to add on "but the apartment is hot", just add on an additional *[Tex \LARGE \wedge] and r to get the final answer of 


*[Tex \LARGE \left(\sim p \wedge q\right) \wedge r]


Take note of the parenthesis. They are for grouping purposes.


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# 2


First, start with a blank table with headers of p, q, ~p, ~p <-> q, ~(~p <-> q)


<table border="1"><th>p</th><th>q</th><th>~p</th><th>~p <-> q</th><th>~(~p <-> q)</th><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td></td></tr></table>



Now fill in the values T, T, F, F for the first column p and the values T, F, T, F for the second column q. This will list all of the possible combos of p and q



<table border="1"><th>p</th><th>q</th><th>~p</th><th>~p <-> q</th><th>~(~p <-> q)</th><tr><td>T</td><td>T</td><td></td><td></td><td></td></tr><tr><td>T</td><td>F</td><td></td><td></td><td></td></tr><tr><td>F</td><td>T</td><td></td><td></td><td></td></tr><tr><td>F</td><td>F</td><td></td><td></td><td></td></tr>
</table>


Negate the first column p to get the third column ~p. In other words, change each T to F (or vice versa) to get the third column ~p

<table border="1"><th>p</th><th>q</th><th>~p</th><th>~p <-> q</th><th>~(~p <-> q)</th><tr><td>T</td><td>T</td><td>F</td><td></td><td></td></tr><tr><td>T</td><td>F</td><td>F</td><td></td><td></td></tr><tr><td>F</td><td>T</td><td>T</td><td></td><td></td></tr><tr><td>F</td><td>F</td><td>T</td><td></td><td></td></tr>
</table>


Recall that p <-> q is only true when p and q have the same truth values (ie when they are equivalent). Use this fact to fill in the 4th column.


<table border="1"><th>p</th><th>q</th><th>~p</th><th>~p <-> q</th><th>~(~p <-> q)</th><tr><td>T</td><td>T</td><td>F</td><td>F</td><td></td></tr><tr><td>T</td><td>F</td><td>F</td><td>T</td><td></td></tr><tr><td>F</td><td>T</td><td>T</td><td>T</td><td></td></tr><tr><td>F</td><td>F</td><td>T</td><td>F</td><td></td></tr></table>



Finally, negate the 4th column to get the fifth and final column.


<table border="1"><th>p</th><th>q</th><th>~p</th><th>~p <-> q</th><th>~(~p <-> q)</th><tr><td>T</td><td>T</td><td>F</td><td>F</td><td>T</td></tr><tr><td>T</td><td>F</td><td>F</td><td>T</td><td>F</td></tr><tr><td>F</td><td>T</td><td>T</td><td>T</td><td>F</td></tr><tr><td>F</td><td>F</td><td>T</td><td>F</td><td>T</td></tr></table>