Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

discrete math:(attached)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
is this right?
1 Attachment
are you proving a tautology?
ya

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

p bar means negation p?
It is not a tautology, use a->b <=> ~a or b and watch out for order of operations.
@Joseph91 , yes @mathmate , it is given that it is tautology, and you must prove it with equilence
see... you just need to PROVE it, .. but my teacher doesnt want the table..
1 Attachment
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
Do you want me to show all the work?
ya please
~p^(p or q) ->q ~p^p or ~p^q -> q
~p^(p or q) ->q ~p^p or ~p^q -> q (distributive property) ok so far?
yup
~p^p = F, and F or anything is anything, so ~p^q -> q ok so far @liliy ?
@liliy still there?
ya
p implication q is equivalent to -pvq -[-p^(pvq)]vq -[-p^pv-p^q]vq -[Fv-p^q]vq -[-p^q]vq pvT=T
woah. wait.. let me digest that
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.
got it! thanks
You're welcome! :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question