Question

Construct a truth table for this statement

q ∨ (p → ∼r)

Answer 1

Can any1 help me with this?

Answer 2

sure

Answer 3

do you know how to start?

Answer 4

What does this sign mean ∨

Answer 5

I'm not familiar with the proper notation

Answer 6

Satellite, I am not sure how to start

Answer 7

you need all possible combinations of T and F for the three variables p, q, r

so it will have 8 rows

Answer 8

Okay

Answer 9

V= OR

It means As long as one side of the variables is true then the whole thing is true

Answer 10

it should look like this
\[\begin{array}{|c|c|c}
P
& Q
& R \\
\hline
T & T & T \\
T & T & F \\
T & F & T \\
T & F & F \\
\hline
F & T & T \\
F & T & F \\
F & F & T\\
F & F & F \\
\hline
\end{array}\]

Answer 11

now we need to add \(\lnot r\)

Answer 12

That is only the first part?

Answer 13

we make no assumptions, we just compute like donkeys
\[\begin{array}{|c|c|c|c}
P
& Q
& R
& \lnot{}R \\
\hline
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 \\
\hline
1 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\hline
\end{array}\]

Answer 14

oh yes, lots more to go

Answer 15

Alright

Answer 16

damn i used 0 and 1, you need T and F
i will let you adjust

Answer 17

I am still working on table one, you can keep going

Answer 18

now we need \(p\to \lnot r\)

Answer 19

for this we look at the column \(p\) and \(\lnot r\) and the implication is only false is \(p\) is true and \(\lnot r\) is false

Answer 20

So O is F and 1 is T?

Answer 21

\[\begin{array}{|c|c|c|c|c}
P
& Q
& R
& \lnot{}R
& P\Rightarrow{}\lnot{}R \\
\hline
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 1 & 1 \\
0 & 1 & 1 & 0 & 1 \\
\hline
1 & 0 & 0 & 1 & 1 \\
1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 & 0 \\
\hline
\end{array}\]

Answer 22

no, 0 are true, 1 are false

Answer 23

See I was right first half is True

Answer 24

1 is True
0 is False

Answer 25

Okay thanks

Answer 26

and finally, we need \(q\lor (p\to \lnot r)\) which, being an or statement, is true unless both are false

Answer 27

\[\begin{array}{|c|c|c|c|c|c}
P
& Q
& R
& \lnot{}R
& P\Rightarrow{}\lnot{}R
& Q\lor{}(P\Rightarrow{}\lnot{}R) \\
\hline
0 & 0 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 & 1 & 1 \\
\hline
1 & 0 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 0 & 0 & 0 \\
1 & 1 & 0 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 0 & 1 \\
\hline
\end{array}\]

Answer 28

Okay this will take a while for me to do, thanks