I found on the internet this rule (A + B = A + A̅ B) and its called Redundant Literal Rule. this is it's OR form. But I want to know is this really form of the Redundant Literal Rule?. I saw in a website there are 2 Types in this Law. What do you think about that. Is that true? (proves can add as comments)
How to Solve The Redundant Literal Rule (OR)?
-1
$\begingroup$
boolean-algebra
-
1What is $A, B$? What is $+$? What is $\overline{A}$? In what "space" all of that happens? There's not enough info to understand the question. – 2017-01-31
-
0I do not believe that the algebraic topology tag is appropriate for this question. – 2017-01-31
2 Answers
2
We have that $$A + A'B = A (1) + A'B = A (1 + B) + A'B $$ $$= A + AB + A'B = A + B (A + A')$$ $$= A+B (1) = A + B $$ where $A'$ is the complement of $A $. Hope it helps.
1
A + A̅B
= (A + AB) + A̅B (Rule : A = A + AB)
= A + (AB + A̅B)
Taking B common,
= A + B (A +A̅)
= A + B (Rule : A + A̅ = 1)
-
0Mine pleasure :-) – 2017-02-01