Question

: Boolean Algebra: How is (Bnot) + (Anot)B equivalent to Bnot + Anot and A(Bnot) + Anot? (Anot is "not A", + is "or" and AB is "A and B")

Answer 1

p.s. is there a way to manipulate the equation to see if they're equivalent? i don't have time to write the whole truth table out for the exam.

Answer 2

According to what you wrote, the second term is A'AB' (using ' for NOT). A'A = AA' = False (something and its opposite can't both be true). Then the problem reduces to B' + A'B = B' + A'. You can verify that with a truth table.

Answer 3

okay, so here's basically what you wrote:

let's suppose F = B' + A'B

F = B' + A'B
F = B' + A'AB' (did B become AB'? if so, how?)
F = B' + A' (A'A is false? i don't quite get it, but even so, what does that have to do with A'AB' becoming just A' ?)

I'm sorry, I'm really new to boolean algebra, it's like a whole new world from what kind of math I've been dealing all my life. Though logic itself is not new, of course, it's just kind of confusing to me haha.

Answer 4

Just for clarification too, I'm asking how is F = B' + A'B equivalent to F = B' + A' and also to F = AB' + A' .

Answer 5

Okay. I totally did not get that from your question. I thought the "and" in the middle was a boolean "and". Ignore what I wrote.

Show these three are equivalent:
1. A'+B'
2. A'B+B'
3. A'+AB'
First realize that 1 and 2 give you 3, or 1 and 3 give you 2. Just swap A and B. So if we can show that 2 implies 1 then we'll be done.

Starting with A'B + B' use De Morgan's. De Morgan's says that (X+Y)' = X'Y'. Taking the complement of both sides gives (X+Y) = (X'Y')'. You'll have to apply De Morgan's in a couple more places. Then you should be able to work that around to get A'+B'. Practice using De Morgan's. It comes in very handy, and it can be confusing when you start taking complements of complements of conjunctions. Just write out every step and check your work.

If you get stuck, post your work.