SOLUTION: Hello,
Can someone please help me I have asked several question but I have not received any answer, if someone would please help me with the following.
a.r: The apartment i
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Question 260642: Hello,
Can someone please help me I have asked several question but I have not received any answer, if someone would please help me with the following.
a.r: The apartment is hot. q: The air conditioner is working. p: The temperature is 90.
Write the following in symbolic form
The temperature is not 90 and the air conditioner is working, but the apartment is hot.
COnstruct a truth table for the statement
~ (~ p<---> q)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
# 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
like so:
Remember that
means "and"
Finally to add on "but the apartment is hot", just add on an additional
and r to get the final answer of
Take note of the parenthesis. They are for grouping purposes.
======================================================================
# 2
First, start with a blank table with headers of p, q, ~p, ~p <-> q, ~(~p <-> q)
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
p | q | ~p | ~p <-> q | ~(~p <-> q) | T | T | | | |
T | F | | | |
F | T | | | |
F | F | | | |
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
p | q | ~p | ~p <-> q | ~(~p <-> q) | T | T | F | | |
T | F | F | | |
F | T | T | | |
F | F | T | | |
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.
p | q | ~p | ~p <-> q | ~(~p <-> q) | T | T | F | F | |
T | F | F | T | |
F | T | T | T | |
F | F | T | F | |
Finally, negate the 4th column to get the fifth and final column.
p | q | ~p | ~p <-> q | ~(~p <-> q) | T | T | F | F | T |
T | F | F | T | F |
F | T | T | T | F |
F | F | T | F | T |
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