Read the two statements shown below. If the weather is cold, Zakir will not go swimming. The weather is not cold, or Zakir will not go swimming. Create truth tables for the logical form of the two statements (not to be submitted). Use the truth tables to determine whether the two statements are logically equivalent. Justify your answer.
let, p : the weather is cold q : Zakir will not go swimming
then, first statement can be written as: p->q second statement is : ~p v q
right ?
Yes. Can you draw the table please ?
sure... but id like you to try.. let me correct if u go wrong... ? do you know how to construct truth table ?
thats what i was going to do, i just need you to set it up
p q p->q ~p v q T T X X T F X X F T X X F F X X
thats the template for any truth table. can u try filling the columns for p->q and ~p->q
what does " -> " mean ?
thats symbol means "implies" its a conditional : "if, then"
and v ?
basic table for \(p\to q\) \[\begin{array}{c|c|c} p & q & p\to q \\ \hline t & t & t \\ t & f & f \\ f & t & t \\ f & t & t \\ \end{array}\]
should memorize this one only false if p is true and q is false
@NyKole are you familiar with the symbols \[p\to q\] \[p \land q\] \[p\lor q\] \[\lnot p\] etc?
no
then it will not be possible to do this problem you need to know the symbols and the truth tables for each
for the p _> would it be T F T F And ~ p v q would be T F F F ???
p q p->q ~p v q T T T T T F F F F T T F F F F F
a truth table is a table, not one column i wrote the truth table for \(p\to q\) meaning "if p then q" or "p implies q" above
close but you have a mistake in the last line of \(p\to q\)
p q p->q ~p v q T T T T T F F F F T T F F F T F
ok good, now the \(p\to q\) part is correct
the \(\lnot p \lor q\) part is wrong we need another step, we need to look at \(\lnot p\)
So that whole column is wrong ?
\[\begin{array}{|c|c|c|c} p & q & \lnot{}p & p\to q \\ \hline t & t & f & t \\ t & f & f & t \\ f & t & t &f \\ f & f & t & t \\ \hline \end{array}\]
yes that column is wrong for the last column \(\lnot p\lor q\) put a T if one or the other or both of \(\lnot p\) or \(q\) is true
for that one look only at the two columns \(\lnot p\) and \(q\) if you see a T in either of those columns, put a T if both are F, put an F i can write it for you if you like
would it be T F T T
\[\begin{array}{|c|c|c|c|c} p & q & \lnot{}p & p\to q & \lnot{}p\lor{}q \\ \hline T & T & F & T & T \\ T & F & F & F & F \\ F & T & T & T & T \\ F & F & T& T& T \\ \hline \end{array}\]
yes you have it!
and since both those columns are identical the statements are logically equivalent
Ok yay. lol now for the second truth table for the question can you help me set it up also pease ?
which is the second one? we have them both in one table
for the second statement? how do you know they are logically equivalent ?
we have a column for \(p\to q\) and also a column for \(\lnot p \lor q\) so we are done
look at the column \(p\to q\) in the truth table and the column \(\lnot p \lor q\) we see that they are identical right?
yes. ok so like how would i justify they are logically equivalent for the question now ? like idk how i would explain it
no reason to make two completely different tables you need only to compare the columns and see if they are the same or not if they are the same, then the statements are equivalent in other words, it is always true that \(p\to q\equiv \lnot p\lor q\)
i would explain it by saying "the column for \(p\to q\) and \(\lnot p\lor q\) are the same therefore the statements are equivalent
it has to be at least 4 to 5 sentences. lol
and don't forget that although this is "logic" it should also make sense so if you think about the statement "if it is raining, i will get wet" and the statement "it is not raining or i will get wet" are exactly the same
you can make up the sentences for yourself i like being brief
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