Question

Does anyone know how to construct a truth table with this? (1) ~q -> p (2) (p^ ~q) ↔ q

Answer 1

these are 2 seperate problems

Answer 2

I have this for the 1st set up... not sure it's ir right.

~q>p
~q p ~q>p

Answer 3

I have this one as the 2nd set up

(p^~q) <>p
P q ~q (p^~q) (p^~q) <>p

Answer 4

lol nevermind

Answer 5

(1) ~p -> Q
the value of p can be only 1 or zero therefore:
if p = 1, then ~p will be 0
if p=0, the ~p will be 1.
regarding the statement ~p->q 'if ~p then q' we could take this to mean:
when p=1 then ~p will be 0 and q=1
when p=0 the ~p will be 1 and q=0
so write down ~p and q as your headers, then 2 boxes under p and q and fill in values as above.

Answer 6

the second one means that it is Q if and only if p and ~q are both true.
in this case because there are two operators there will be 4 combinations:
so draw your box as before with 4 headers, p, ~q, p^~q and q
underneath each of these you will have 4 boxes.
under the p box write: 0,1,0,1, and under the ~q write 1,1, 0, 0 .
that gives you all possible combinations between these two operators.
Now you just need to work out the value of p^~q and fill those in in the next set of boxes.
p ^ ~q
0 (0 and, not q where q=1 is 1 and 0 which is 0 and 0 )
1 (1 and, not q where q =1 is 1 and not 0 which is 1 and 1)
0 (0 and, not q where q=0 is 0 and not 0 which is 0 and 1)
1 (1 and, not q where q=0 is 1 and not 0 which is 1 and 1)

So p^~q <--> q for value in the second and last row. otherwise 0 (or false).

Answer 7

you might also like to check out Demorgans laws for simplifying a lot of these statements as they get more complex.

Answer 8

BPDlkeme, thanks a million!!!!!!

Answer 9

good night