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Mathematics 22 Online
OpenStudy (anonymous):

ATT WASIQSS: Given any statement, is it possible to find a logically equivalent statement using only ~ and ⋀ ? Justify your answer.

OpenStudy (wasiqss):

omg frikkin difficult :P

OpenStudy (anonymous):

completeness of boolean algebra

OpenStudy (wasiqss):

Well pippa no idea :P

OpenStudy (inkyvoyd):

Tell me what those two signs mean, and let me thinky thinky?

OpenStudy (anonymous):

i wld say no

OpenStudy (anonymous):

~ negation ^ and

OpenStudy (inkyvoyd):

Yes, it is

OpenStudy (inkyvoyd):

Well, I'm not sure, but I know that any logic gate can be constructed with only Nand gates

OpenStudy (anonymous):

yeah i think so too, but it is rather a pain

OpenStudy (inkyvoyd):

I've done that part myself.

OpenStudy (wasiqss):

Im running :D

OpenStudy (anonymous):

hmmm never seen that i have only learnt abt de morgans laws so far

OpenStudy (anonymous):

what is nand gates

OpenStudy (anonymous):

it is electronics

OpenStudy (inkyvoyd):

in o 00 1 11 0 01 1 10 1

OpenStudy (anonymous):

accomplish inversion and and

OpenStudy (inkyvoyd):

Basically the negation of the AND gate

OpenStudy (anonymous):

i think the idea is this first you can show that you can use \(\lnot, \land \) and \(\lor\) to create all "disjunctive normal forms"

OpenStudy (anonymous):

but i can only use ^ and ~

OpenStudy (inkyvoyd):

Well, a combination of AND and NOT can create the others.

OpenStudy (anonymous):

then by demorgan, \(p\lor q\) is equavelent to \(\lnot(\lnot p \land \lnot q)\)

OpenStudy (anonymous):

so that gets rid of the \(\lor\) and you are done, modulo a raft of details

OpenStudy (anonymous):

ohh i misread teh question hahahahahhaha

OpenStudy (anonymous):

i am dumb wasiqss lol

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