## liliy 3 years ago discrete math:(attached)

1. liliy

is this right?

2. josiahh

are you proving a tautology?

3. liliy

ya

4. Joseph91

p bar means negation p?

5. mathmate

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

6. liliy

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

7. liliy

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

8. mathmate

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

9. mathmate

Do you want me to show all the work?

10. liliy

11. mathmate

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

12. mathmate

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

13. liliy

yup

14. mathmate

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

15. mathmate

@liliy still there?

16. liliy

ya

17. Joseph91

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

18. liliy

woah. wait.. let me digest that

19. mathmate

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.

20. liliy

got it! thanks

21. mathmate

You're welcome! :)