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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

we can do this

- anonymous

if you are here that is, it is not that hard

- anonymous

ready?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

yes

- anonymous

I have no idea how to do this lol @satellite73

- anonymous

ok first we start here
\[\begin{array}{c|c}
p
& q \\
\hline
& \\
& \\
& \\
& \\
\end{array}\]

- 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

that that each possible pair is there
have you seen something like this before?

- 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

\(p\to q\) 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

how we doing so far? are you totally lost, or is it okay ?

- anonymous

oh okay im starting to get it

- anonymous

the first two columns are there to get all possible combinations of \(T\) and\(F\) for \(p,q\)

- anonymous

the third column \(p\to q\) as i said is always true unless \(p\) is \(T\) and \(q\) is \(F\)

- anonymous

as your read across
two more to go

- 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

\[\begin{array}{|c|c|c|c|}
p
& q
& p\to q
&p\land (p\to q) \\
\hline
T & T &T & \\
T & F & F&\\
F & T & T&\\
F & F & T&\\
\hline
\end{array}\]

- anonymous

we will get there if you have the patience for it
we are not there yet

- anonymous

we have to fill in the column \(p\land (p\to q)\)

- 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

\[\begin{array}{|c|c|c|c|}
p
& q
& p\to q
&p\land (p\to q) \\
\hline
T & T &T &T \\
T & F & F&F\\
F & T & T&F\\
F & F & T&F\\
\hline
\end{array}\]

- anonymous

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

What does the ^ mean

- anonymous

\(p\land q\) means "\(p\) AND \(q\)"
it is only True if both \(p\) and \(q\) are True

- anonymous

like saying "it is raining and cold" is only true if it is both raining and cold

- anonymous

okay but i dont get what the last column means all together. p ∧ (p → q)

- anonymous

in english it is kind of funny
it says "p and p implies q"

- anonymous

oh okay

- 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

oh okay!

- anonymous

last column ready? it is going to take me a minute to format it

- anonymous

okay ready

- 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

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

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

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

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

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

yeah so they are equivalent?

- 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

oh i thought were on 4 lol

- anonymous

the answer to that one is "yes, it is a tautology"

- anonymous

#4 requires doing #2, which is a lot easier than the last one

- anonymous

okay how do i figure it out

- anonymous

ok here we go

- 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

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

do you know what \(\lnot p\) means?

- anonymous

isnt it just the opposite of p?

- 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

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

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

\[\begin{array}{|c|c|c|c|c|c}
p
& q
& \lnot{}p
& \lnot{}q
& p\to q
& \lnot{}q\to\lnot{}p \\
\hline
T & T & F &F &T \\
T & F &F & T&F & \\
F & T & T& F & T& \\
F & F & T& T& T \\
\hline
\end{array}\]

- anonymous

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

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

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

now we look at the last two columns and see that they are identical
guess what that means?

- anonymous

they are equivalent!

- 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

Oh okay could you help me set up the chart?

- anonymous

@satellite73

- anonymous

set up what chart?

- anonymous

Oh nvm you already did i think for number 2 it says to set up the truth table

- anonymous

lordamercy
what do you think we were doing for the past hour?!

- anonymous

lol but which chart is the one that proves they are equivalent

- 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

\[\begin{array}{|c|c|c|c|c|c}
p
& q
& \lnot{}p
& \lnot{}q
& p\to q
& \lnot{}q\to\lnot{}p \\
\hline
T & T & F &F &T&T \\
T & F &F & T&F &F \\
F & T & T& F & T& T \\
F & F & T& T& T &T\\
\hline
\end{array}\]this is the truth table for both \(p\to q\) and \(\lnot q\to \lnot p\)
under both, you see the same thing

Looking for something else?

Not the answer you are looking for? Search for more explanations.