Confusing about logic gates

Question

Says i have this logic :

X = (A & B) | ~B

Which can be shorten to :

X = ~(~A & B)

and then :

X = A | ~B

so :

(A & B) | ~B = A | ~B

About this one, i can prove it drawing a truth table, but i still can not shorten the logic. The guy gave me the shorten logic told me to use De Morgan, but I can't do it through i'm aware of basic De Morgan law

Likewise :

A AND (A OR B) <=> A

This one i have no idea how it can be shorten like that!

Any explanation is greatly appreciated !!!

@DanZimm sorry, edited, I can prove it's if i draw a truth table, what'm I asking is how can i shorten the logics like that ? — f855a864, Aug 15 '14 at 09:27

score 4 · Accepted Answer · edited Apr 23 '15 at 06:23

4

HINT

AND distributes over OR : $A~\&~(B|C) = (A\&B)~|~(A\&C) $

OR distributes over AND : $A~|~(B\&C) = (A|B)~\&~(A|C)$

Your last example : $A~ \& ~(A | B) = (A|0)~\&~(A|B) = A~|~(0\&B) = A~|~0 = A$

Refer this for a list of useful Boolean algebra laws/theorems/proofs

edited Apr 23 '15 at 06:23

answered Aug 15 '14 at 09:44

AgentS

12,195

thanks, how about the first example ? – f855a864 Aug 15 '14 at 09:53
simply distribute OR : $X = (A&B)~|~\neg B = (A|\neg B)~&~(B|\neg B ) = (A|\neg B)~&~(1 ) =A|\neg B$ – AgentS Aug 15 '14 at 09:59
sorry for the late reply, i'll check it out later – f855a864 Aug 15 '14 at 13:22
I checked it out, the list is very useful, i'm gonna memorize that, thanks a lot mate – f855a864 Aug 15 '14 at 13:36
you're welcome :) – AgentS Aug 15 '14 at 13:43

Confusing about logic gates

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