Psymon
  • Psymon
Construct a truth table to show that the conditional is equivalent to ~p v q.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Psymon
  • Psymon
I guess I don't fully understand the truth values of the terms well enough to fill these out properly. Too much uncertainty on my part, so hoping someone can explain, lol.
anonymous
  • anonymous
do you know the truth table for \(p\to q\)?
Psymon
  • Psymon
Not sure. I can take a guess but Im not positive about it.

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anonymous
  • anonymous
the first step is to write all combinations of T and F for \(p\) and \(q\) usually it is written line this \[\begin{array}{|c|c|c} p & q & p\to q \\ \hline T & T & \\ T & F & \\ F & T & \\ F & F & \\ \hline \end{array}\]
anonymous
  • anonymous
so in each row, there is a combination of T and F for example in the second row, \(p\) is True and \(q\) is False
Psymon
  • Psymon
\[p \rightarrow q\] |dw:1378394832502:dw| Im pretty sure I remember false implying true but I never understood why. Although I could be confusing that with another implication, I forget
anonymous
  • anonymous
that one is True
anonymous
  • anonymous
for example, i says "if it is raining, then i carry an umbrella" suppose it is not raining, i.e. p is false then i can still carry an umbrella. the statement \(p\to q\) doesn't say anything about what happens if \(p\) is false. \(q\) could be true, or \(q\) you be false, but \(p\to q\) is still true
anonymous
  • anonymous
truth table looks like \[\begin{array}{|c|c|c} p & q & p\to q \\ \hline T & T & T \\ T & F & F\\ F & T & T\\ F & F & T\\ \hline \end{array}\]
Psymon
  • Psymon
So....if P then Q is ONLY in reference to P being true? Like, P being false cannot imply anything because the initial statement is only regards to P being true?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i can come up with a math example if you like, but i like my rain example
anonymous
  • anonymous
if it is saturday, i go to the movies it is not saturday. i can still go to the movies right?
Psymon
  • Psymon
Alright, that makes sense then. So then having ~p v q. Is that at all dependent on the implication or is it always independent? And okay, so simply based on what is possible given the conditions of the implication.
anonymous
  • anonymous
write down the truth table for \(\lnot p\lor q\) and you will see that the column is identical to the column for \(p\to q\) that means by definition that they are equivalent
Psymon
  • Psymon
Regardless of what P and Q actually are, right?
anonymous
  • anonymous
using all combinations of T and F as in the example i showed
Psymon
  • Psymon
|dw:1378395407108:dw| wait, do I still use ~p in the column and q int he second or do I need to only use p and q?
anonymous
  • anonymous
again start like this \[\begin{array}{|c|c|c|c} p & q & \lnot{}p & \lnot{}p\lor{}q \\ \hline T & T & & \\ T & F & & \\ F & T & & \\ F & F & & \\ \hline \end{array}\]
anonymous
  • anonymous
you see the first two columns generate all possible combinations of T and F for p and q
anonymous
  • anonymous
that is how you always start these if there is p q and r you need 8 rows
Psymon
  • Psymon
So if I were to compare say the inverse and the converse, would I still always start with the first two columns you have?
anonymous
  • anonymous
yes inverse, converse, \(\lnot q\lor (p\land q)\) whatever if you have two statements, you have to have 4 rows to give all combinations of T and F
anonymous
  • anonymous
if you have \(n\) statements you need \(2^n\) rows kind of gets messy
Psymon
  • Psymon
Gotcha. |dw:1378395767757:dw| Okay, I think Im getting this then, thanks ^_^
anonymous
  • anonymous
yeah that is exactly right the fact that the column under \(\lnot p\lor q\) is the same as for \(p\to q\) tells you that they are logically equivalent but so does common sense if it is saturday, i go to the movies it is not saturday, or i go to the movies. same thing
Psymon
  • Psymon
Right. Alright, cool. Yeah, just knowing how to start it off helped out a lot, too. Thanks again : )

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