Question

discrete math:(attached)

Answer 1

is this right?

Answer 2

are you proving a tautology?

Answer 3

ya

Answer 4

p bar means negation p?

Answer 5

It is not a tautology,
use a->b <=> ~a or b
and watch out for order of operations.

Answer 6

@Joseph91 , yes
@mathmate , it is given that it is tautology, and you must prove it with equilence

Answer 7

see... you just need to PROVE it, .. but my teacher doesnt want the table..

Answer 8

You're right, it is a tautology, I made a mistake in one of the terms.
Still,
start with the distributive property:
~a^(a or b) <=> ~a^a or ~a^b
then use
~a and ~b <=> F
and then
~a or b <=> b
That leaves you with
~p^q -> q
and you'll get it after that using
a->b <=> ~a or b

Answer 9

Do you want me to show all the work?

Answer 10

ya please

Answer 11

~p^(p or q) ->q
~p^p or ~p^q -> q

Answer 12

~p^(p or q) ->q
~p^p or ~p^q -> q (distributive property)
ok so far?

Answer 13

yup

Answer 14

~p^p = F, and F or anything is anything, so
~p^q -> q
ok so far @liliy ?

Answer 15

@liliy still there?

Answer 16

ya

Answer 17

p implication q is equivalent to -pvq
-[-p^(pvq)]vq
-[-p^pv-p^q]vq
-[Fv-p^q]vq
-[-p^q]vq
pvT=T

Answer 18

woah. wait.. let me digest that

Answer 19

So Joseph91 has completed the proof.
The last step T is from ~q or q = T
after substituting ~(~p^q) by p or ~q due to de morgan's theorem.

Answer 20

got it! thanks

Answer 21

You're welcome! :)