anonymous
  • anonymous
construct and complete a truth table with the following headings: 1) p 2) q 3) p → q 4) p ∧ (p → q) 5) [p ∧ (p → q)] → q
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
we can do this
anonymous
  • anonymous
if you are here that is, it is not that hard
anonymous
  • anonymous
ready?

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anonymous
  • anonymous
yes
anonymous
  • anonymous
I have no idea how to do this lol @satellite73
anonymous
  • anonymous
ok first we start here \[\begin{array}{c|c} p & q \\ \hline & \\ & \\ & \\ & \\ \end{array}\]
anonymous
  • anonymous
then we put all combinations of t and f underneath usually like this \[\begin{array}{|c|c} p & q \\ \hline T & T \\ T & F \\ F & T \\ F & F \\ \hline \end{array}\]
anonymous
  • anonymous
that that each possible pair is there have you seen something like this before?
anonymous
  • anonymous
next step is \[\begin{array}{|c|c|c} p & q & p\to q \\ \hline T & T & \\ T & F & \\ F & T & \\ F & F & \\ \hline \end{array}\]
anonymous
  • anonymous
\(p\to q\) is always true, unless \(p\) is true and \(q\) is false it will look like this \[\begin{array}{|c|c|c} p & q & p\to q \\ \hline T & T &T \\ T & F & F\\ F & T & T\\ F & F & T\\ \hline \end{array}\]
anonymous
  • anonymous
how we doing so far? are you totally lost, or is it okay ?
anonymous
  • anonymous
oh okay im starting to get it
anonymous
  • anonymous
the first two columns are there to get all possible combinations of \(T\) and\(F\) for \(p,q\)
anonymous
  • anonymous
the third column \(p\to q\) as i said is always true unless \(p\) is \(T\) and \(q\) is \(F\)
anonymous
  • anonymous
as your read across two more to go
anonymous
  • anonymous
This is what the project says: On a sheet of paper, construct and complete a truth table with the following headings: p q p → q p ∧ (p → q) [p ∧ (p → q)] → q 2. Construct and complete a truth table in order to determine if the following statements are logically equivalent: p → q and ~q → ~p 3. Explain in the box below if the statement [p ∧ (p → q)] → q is a tautology or not. 4. Explain in the box below if the statements p → q and ~q → ~p are logically equivalent or not. 5. Turn your paper with your truth tables in to your teacher.
anonymous
  • anonymous
\[\begin{array}{|c|c|c|c|} p & q & p\to q &p\land (p\to q) \\ \hline T & T &T & \\ T & F & F&\\ F & T & T&\\ F & F & T&\\ \hline \end{array}\]
anonymous
  • anonymous
we will get there if you have the patience for it we are not there yet
anonymous
  • anonymous
we have to fill in the column \(p\land (p\to q)\)
anonymous
  • anonymous
for this column you only look at the column under \(p\) and the column under \(p\to q\) the "and" statement is only true if there is a T in both rows
anonymous
  • anonymous
\[\begin{array}{|c|c|c|c|} p & q & p\to q &p\land (p\to q) \\ \hline T & T &T &T \\ T & F & F&F\\ F & T & T&F\\ F & F & T&F\\ \hline \end{array}\]
anonymous
  • anonymous
if i have totally lost you let me know or if you have a specific question please ask we have one more column to fill
anonymous
  • anonymous
What does the ^ mean
anonymous
  • anonymous
\(p\land q\) means "\(p\) AND \(q\)" it is only True if both \(p\) and \(q\) are True
anonymous
  • anonymous
like saying "it is raining and cold" is only true if it is both raining and cold
anonymous
  • anonymous
okay but i dont get what the last column means all together. p ∧ (p → q)
anonymous
  • anonymous
in english it is kind of funny it says "p and p implies q"
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
if we change p to "it is sunny" and q to "i go to the beach" it says "it is sunny and if it is sunny i will go to the beach"
anonymous
  • anonymous
oh okay!
anonymous
  • anonymous
last column ready? it is going to take me a minute to format it
anonymous
  • anonymous
okay ready
anonymous
  • anonymous
\[\begin{array}{|c|c|c|c|c} p & q & p\to q &p\land (p\to q) &[p\land (p\to q)]\to q \\ \hline T & T &T &T& \\ T & F & F&F&\\ F & T & T&F&\\ F & F & T&F&\\ \hline \end{array}\]
anonymous
  • anonymous
the reason they have you do one at a time is so we can understand the last one the last one is the \([p\land (p\to q)]\to q\) which is always true unless \(p\land (p\to q)\) is true and \(q\) is false so we look at the column under \(p\land (p\to q)\) and \(q\) and if we see \(T, F\) we put \(F\) otherwise we put \(T\)
anonymous
  • anonymous
but if you look carefully, you see that it is never the case if \(p\land (p\to q)]\) is \(T\) then \(q\) is also \(T\) in that row \[\begin{array}{|c|c|c|c|c} p & q & p\to q &p\land (p\to q) &[p\land (p\to q)]\to q \\ \hline T &\color{red} T &T &\color{red}T& \\ T & F & F&F&\\ F & T & T&F&\\ F & F & T&F&\\ \hline \end{array}\]
anonymous
  • anonymous
therefore the last column gets all T\[\begin{array}{|c|c|c|c|c} p & q & p\to q &p\land (p\to q) &[p\land (p\to q)]\to q \\ \hline T & T &T &T& T\\ T & F & F&F&T\\ F & T & T&F&T\\ F & F & T&F&T\\ \hline \end{array}\]
anonymous
  • anonymous
now we can answer question 3 as well since the last column has all \(T\) that means it is always true, and is a "tautology"
anonymous
  • anonymous
back to the previous example, the last column says "it is sunny, and if it is sunny i go to the beach, then i go to the beach" pretty obviously true right?
anonymous
  • anonymous
yeah so they are equivalent?
anonymous
  • anonymous
hold the phone you need two different statements to ask "are they equivalent" we were answering this question 3. Explain in the box below if the statement [p ∧ (p → q)] → q is a tautology or not.
anonymous
  • anonymous
oh i thought were on 4 lol
anonymous
  • anonymous
the answer to that one is "yes, it is a tautology"
anonymous
  • anonymous
#4 requires doing #2, which is a lot easier than the last one
anonymous
  • anonymous
okay how do i figure it out
anonymous
  • anonymous
ok here we go
anonymous
  • anonymous
\[\begin{array}{|c|c|c|c|c|c} p & q & \lnot{}p & \lnot{}q & p\to q & \lnot{}q\to\lnot{}p \\ \hline T & T & & & & \\ T & F & & & & \\ F & T & & & & \\ F & F & & & & \\ \hline \end{array}\]
anonymous
  • anonymous
as before, first two columns give all combinations of T and F now to get \(\lnot p\) and \(\lnot q\) change all T to F and F to T
anonymous
  • anonymous
do you know what \(\lnot p\) means?
anonymous
  • anonymous
isnt it just the opposite of p?
anonymous
  • anonymous
\[\begin{array}{|c|c|c|c|c|c} p & q & \lnot{}p & \lnot{}q & p\to q & \lnot{}q\to\lnot{}p \\ \hline T & T & F &F \\ T & F &F & T& & \\ F & T & T& F & & \\ F & F & T& T& \\ \hline \end{array}\] yes it means "not p"
anonymous
  • anonymous
you see how i got the column \(\lnot p\)? i looked at the column for \(p\) and if i see T i put F and if i see F i put T same for \(\lnot q\)
anonymous
  • anonymous
now we already did \(p\to q\) for the last question it is always T unless \(p\) is T and \(q\) is F, in the second row
anonymous
  • anonymous
\[\begin{array}{|c|c|c|c|c|c} p & q & \lnot{}p & \lnot{}q & p\to q & \lnot{}q\to\lnot{}p \\ \hline T & T & F &F &T \\ T & F &F & T&F & \\ F & T & T& F & T& \\ F & F & T& T& T \\ \hline \end{array}\]
anonymous
  • anonymous
and finally we look at \(\lnot q\to \lnot p\) which is always true unless \(\lnot q\) it T and \(\lnot p\) is F
anonymous
  • anonymous
that only occurs here \[\begin{array}{|c|c|c|c|c|c} p & q & \lnot{}p & \lnot{}q & p\to q & \lnot{}q\to\lnot{}p \\ \hline T & T & F &F &T \\ T & F &\color{red}F & \color{red}T&F & \\ F & T & T& F & T& \\ F & F & T& T& T \\ \hline \end{array}\]
anonymous
  • anonymous
fill it in like this \[\begin{array}{|c|c|c|c|c|c} p & q & \lnot{}p & \lnot{}q & p\to q & \lnot{}q\to\lnot{}p \\ \hline T & T & F &F &T&T \\ T & F &\color{red}F & \color{red}T&F &F \\ F & T & T& F & T& T \\ F & F & T& T& T &T\\ \hline \end{array}\]
anonymous
  • anonymous
now we look at the last two columns and see that they are identical guess what that means?
anonymous
  • anonymous
they are equivalent!
anonymous
  • anonymous
exactly back to our english example if it is sunny, i will go to the beach if i did not go to the beach, then it is not sunny they say the same thing
anonymous
  • anonymous
Oh okay could you help me set up the chart?
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
set up what chart?
anonymous
  • anonymous
Oh nvm you already did i think for number 2 it says to set up the truth table
anonymous
  • anonymous
lordamercy what do you think we were doing for the past hour?!
anonymous
  • anonymous
lol but which chart is the one that proves they are equivalent
anonymous
  • anonymous
the entire table is the truth table for both the fact that the last two columns are identical is what tells you the two statements are equivalent
anonymous
  • anonymous
\[\begin{array}{|c|c|c|c|c|c} p & q & \lnot{}p & \lnot{}q & p\to q & \lnot{}q\to\lnot{}p \\ \hline T & T & F &F &T&T \\ T & F &F & T&F &F \\ F & T & T& F & T& T \\ F & F & T& T& T &T\\ \hline \end{array}\]this is the truth table for both \(p\to q\) and \(\lnot q\to \lnot p\) under both, you see the same thing

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