Question

make a truth table for this problem. :) (~p V q) --> p

Answer 1

i am trying to work this out .. :)

Answer 2

What've you got so far? Have you done the p, ~p, q columns yet?

Answer 3

p -p q -p^q
t f f f
f t f f
f t f
t t t

Answer 4

so if ~ p Or q is T then its T right?

Answer 5

.... mixed and with or .... dint i

Answer 6

Right. Amistre got his operator mixed up.

Answer 7

At least I'm not the only one.

Answer 8

p -p q -porq
t f f f
f t f t
f t t
t t t

:)

Answer 9

if i recall:

p -> q
t t t
t f f
f t t
f f t

is the basic set up for the conditional

Answer 10

for the ~p V q is t f t f t t t

Answer 11

should it be 8 rows down right?

Answer 12

-pVq -> p
f t = t
t t = t
t t = t
t t = t
f f = t
t f = f
t f = f
t f = f

Answer 13

i think 8 is good for rows

Answer 14

The second one should be f

Answer 15

if T then T would be T, right?

Answer 16

right but you have it wrong... the 2nd row should be T F F and then 4 column should be F

Answer 17

am i wrong polnak?

Answer 18

Sorry, lemme check here..

Answer 19

do i have to have column for -->p also?

Answer 20

You need a column for \[(\lnot p \vee q) \implies p\]

Answer 21

so i should have 5 columns right?

Answer 22

Yes

Answer 23

(~pV q)-->p is my final column

Answer 24

Correct

Answer 25

this is what i have to column 4 ... t f t f t t t f

Answer 26

\[\begin{array}{|c|c|c|c|c}
p & \lnot p & q & \lnot p \vee q & (\lnot p \vee q ) \implies p\
\\ \hline T & F & T & T
\\ T & F & F & F
\\ F & T & T & T
\\ F & T & F & T
\end{array}\]

Answer 27

I'm not sure why you have so many rows.

Answer 28

(~pV q)-->p is my final column

Answer 29

idk thats what was told by our professor.. lol

Answer 30

There are only 2 variables, so there should only be \(2^2= 4\) rows

Answer 31

If there were 3 variables (p, q, r) for example then there would be \(2^3 = 8\) rows

Answer 32

but we do have 3 of them ~ p p and q

Answer 33

no, ~p depends on p it is not independent.

Answer 34

oh ok...

Answer 35

the number of rows depends on the number of independent variables.

Answer 36

but we have ~ p p and q for the column

Answer 37

Yes, but again, p and q can vary freely. ~p is ALWAYS the opposite of p.

Answer 38

so ~p depends on the value of p it is not a free variable

Answer 39

ok.. so column 4 should be
t f t f

Answer 40

No, lets see your table so far and see what's goin on.

Answer 41

p q ~p (~p V q) (~pVq) -->p
T T F T
T F F F
F T T T
F F T F

Answer 42

the 8 rows i believe comes from introducing the "->p" part, to accomodate p as t and f. a bit redundant to me tho

Answer 43

That last row in the fourth column should be True. ~p is True, so ~p OR q is True because at least ~p is True.

Answer 44

ok

Answer 45

Amistre, you only have rows for the different combinations of your independent variables. Any derived values don't need their own rows as they cannot vary independently of the independent variables.

Answer 46

owowowow ... i read :a brand new world" and "man search for meaning" this week so I had to forgot a few things :)

Answer 47

Heh.. I know how that goes.

Answer 48

so the final column would be
T F F F

Answer 49

So any time \(\lnot p \vee q\) is F, \((\lnot p \vee q)\implies p\) will be T (regardless of p)

Answer 50

But if \(\lnot p \vee q\) is T then \((\lnot p \vee q) \implies p\) is T ONLY IF p is T

Answer 51

If the antecedent is True then the truth of the implication depends on the consequent. But if the antecedent is False, it doesn't matter what the consequent is. The implication is vacuously True.

Answer 52

So it's T, T, F, T

Answer 53

i thought there can only be 1 T for the final column?

Answer 54

Nope, that's what I was saying.. If Column 4 is F then no matter what p is, column 5 will be T.

Answer 55

If column 4 is T, then column 5 will be T only if p is T.

Answer 56

This is the general form of an implication:
\[\begin{array}{|cc|c} g&y& g\implies y\\\hline T & T & T\\ T & F & F\\ F & T & T \\ F & F & T \end{array}\]

Answer 57

i must have column 4 wrong.. i have T F T T

Answer 58

that is right for column 4

Answer 59

Whoopse the answer I gave before is wrong.

\[\begin{array}{|c|c|c|c|c}
p & \lnot p & q & \lnot p \vee q & (\lnot p \vee q ) \implies p\
\\ \hline T & F & T & T &T
\\ T & F & F & F & T
\\ F & T & T & T & F
\\ F & T & F & T & F
\end{array}\]

Answer 60

I have a hard time unless I do it in the table ;)

Answer 61

u put the ~ p in a different spot then me.. i put it after the q

Answer 62

The order of the columns won't change anything

Answer 63

oh ok...
i have the same answer as you ;)

Answer 64

\(T \implies T\) is \( T\)
\(F \implies T\) is \( T\)
\(T \implies F\) is \( F\)
\(T \implies F\) is \( F\)

Answer 65

that is for IF right ? -->

Answer 66

Yeah it's like 'if ... then ...'

Answer 67

ok

Answer 68

ty hon... i gotta get ready for work.. i will be back with another truth table tonight.. lol

Answer 69

\[p \implies q\]
if p is True then q Must be True

Answer 70

Ok, have a good day at work =)