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Mathematics 24 Online
OpenStudy (anonymous):

Use De Morgan’s laws to determine whether the two statements are equivalent. ~ (p ∧ q), ~p ∧ ~q

OpenStudy (anonymous):

its been a while since ive dealt with truth tables and things like it, but when you negate something with a ^ or v in it, dont you switch it to the other?

OpenStudy (anonymous):

hold on im not sure let me look at me txt book!

OpenStudy (anonymous):

i might be completely wrong, its been a while.

OpenStudy (anonymous):

i think my book is in my car <.< im to tired to get it lol

OpenStudy (anonymous):

~( p∧q)⇔~ p ∨ ~q by law 1 and this is not equivalent to ~ p ∧ ~ q, since if p is true and q is false, the first statement is true but the second is false. i got this from a website not sure if it correct!

OpenStudy (anonymous):

brb

OpenStudy (anonymous):

i think thats right. I remember always having to switch the sign when you negate.

OpenStudy (anonymous):

ima go get my book lol >.>

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

im back!

OpenStudy (anonymous):

joe how would u write in ur own words?

OpenStudy (anonymous):

if the problem didnt say "using De Morgan's Law", i would just use truth tables to show they arent equivalent.

OpenStudy (anonymous):

but, since it say that, im looking through my book to see which Law would directly oppose that statement

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

alright i got it. i'll post a pic in a sec.

OpenStudy (anonymous):

by definition, the negation of (p ^ q) is (~p OR ~q), so they are not equivalent

OpenStudy (anonymous):

@ Joe i thnk i got it thanks!

OpenStudy (anonymous):

...and it looks like we arent able to post pics for the time being =/ bleh. Well, one of de Morgan's Laws is: \[\lnot(p\land q) \Leftrightarrow (\lnot p)\lor (\lnot q)\]

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