## rajat97 one year ago Mathematical logic prove the following equivalence without using truth table: ~(p v q) v (~p ^ q) = ~p

1. freckles

You can make a truth table.

2. rajat97

it is to be done without a truth table!

3. rajat97

i'm sorry i forgot to add it

4. rajat97

5. freckles

for some reason I'm thinking demorgan's law

6. rajat97

yeah i did it and then tried to apply the distributive law but it was going way too long

7. freckles

$-(p \text{ or } q) \text{ or } ( -p \text{ and } q ) \\ \text {recall } -T \text{ or } -Q \equiv -(T \text{ and } Q) \\ \\ -(p \text{ or } q ) \text{ or } -(p \text{ or } - q) \\$ going to try to use it again $-([p \text{ or } q ]\text{ and } [p \text{ or } -q])$ I guess you can try that distributive law too $-(p \text{ or } (q \text{ and } -q ))$

8. freckles

I think that should make it pretty easy on you now :)

9. freckles

you know since q and not q is definitely what kind of statement?

10. rajat97

that's how it is solved in the textbook but i don't understand how did you use the distributive law it should be −([p or q] and [p or −q]) is equivalent to -[p and (p or -q) or q and (p or -q)] according to me

11. freckles

distributive law says: $(P \text{ or } Q ) \text{ and } (P \text{ or } R) \equiv P \text{ or } ( Q \text{ and } R ) \\ \text{ where we have } \\ P=p \\ Q=q \\ R=-q$ plug in

12. rajat97

i may be a bit irritating but i didn't get that

13. freckles

wait which part? do you understand that is one of the distributive laws?

14. freckles

like and side the not we had: $(p \text{ or } q ) \text{ and } ( p \text{ or } -q)$

15. freckles

this is in that exact form I mentioned above where P=p and Q=q and R=-q

16. freckles

just plug into that law above it

17. freckles

$p \text{ or } (q \text{ and } -q)$

18. freckles

and of course bring down the not sign that was in front of all of that

19. freckles

like inside the not we had*

20. Loser66

To me, it is quite simple not(p or q) AND (not p AND q = (not p AND not q) AND(notp AND q) Now, all of them are AND, we can take off the parentheses $$\neg p \wedge \neg q\wedge \neg p \wedge q\equiv \neg p$$, since $$\neg q \wedge q =0$$

21. rajat97

no i didn't understand (P or Q) and (P or R)≡P or (Q and R) where we have P=pQ=qR=−q according to me it should be [P and (P or R)] or [Q and (P or R)]

22. freckles

according to you?

23. rajat97

yeah i'm sorry i mean according to the thing i read in the textbook

24. freckles

oh I don't know about that rule that what I have to be something else I try to prove but I was using the distributive law which says https://en.wikipedia.org/wiki/Logical_equivalence Instead of typing it out again

25. freckles

also @Loser66 's thing looks good to me he made it a completely an and operation thing

26. freckles

she sorry

27. freckles

oh that is because he changed the original statement

28. freckles

she sorry again @Loser66

29. rajat97

okay i'll try to get it just a moment

30. Loser66

hahaha. it's ok @freckles I worked on his/her original problem (on the very first post)

31. freckles

yeah the statement would have been easier if it was and in the middle

32. ganeshie8

 ~(p v q) v (~p ^ q) (~p ^ ~q) v (~p ^ q) ~p ^ (~q v q) ~p ^ 1 ~p 

33. rajat97

thanks a lot everybody for your help we are allowed to choose only one best answer or else i would give each one of you a medal one last question please what if it is (p ^ q) v (q v p) how would the working with distributive law go?

34. Loser66

=(p v (q v p)) ^ (q v (q v p)) = (p v q) ^ (q v p) = p v q

35. Loser66

=$$q\rightarrow \neg p$$