Laws of Boolean algebra

[dadaqt]

3. Consider the Boolean expression F(A,B,C) = ABC + A'C' + A'B'.

(a) Using the laws of Boolean algebra, derive an expression that uses only AND and NOT operators.
(b) Using the laws of Boolean algebra, derive an expression that uses only OR and NOT operators.

First, I am sorry for my poor grammar. :'(

I have tried every laws I could find and still couldn't solve it, heres 1 example of what I have done:

ABC + A'C' + A'B'
= ABC + A'C'(B + B') + A'B'(C+C')
= ABC + A'C'B + A'B'C' + A'B'C + A'B'C'
= (ABC + A'BC' ) + (ABC + A'B'C) + (ABC + A'B'C') + (ABC + A'B'C')
= B(AC + A'C') + C(AB + A'B') + (1) + (1)
= B + C + 1 + 1

and a few other tries, but the one above is the best I have got. I dont think any of those are right.

Please help.

thanks

[griswolf]

The trick is to to use the rule for distributing negation over a conjunction (OR) or a disjunction (AND): ~(AB) = ~A + ~B and ~(A+B) = ~A~B (I'm using the ~ for the negation operator, where you use an apostrophe. With that and the rule that ~~A == A you have all you need. Here's the first few lines of the derivation to get to a full conjuctive form.

ABC + ~A~C + ~A~B
= ~~(ABC) + ~~(~A~C) + ~~(~A~B)
= ~(~A + ~B + ~C) + ~~(~A~C) + ~~(~A~B)
= ~(~A + ~B + ~C) + ~(~~A + ~~C) + ~~(~A~B)
= ~(~A + ~B + ~C) + ~(A + C) + ~~(~A~B)
(I'm sure you can do the rest)

An exactly mirror technique gets you to the full disjunctive form.

[gooradog]

Griswolf: hi i'm trying to solve this question as well but i'm still not getting it can u explain it to me if possible thank you

[gooradog]

The trick is to to use the rule for distributing negation over a conjunction (OR) or a disjunction (AND): ~(AB) = ~A + ~B and ~(A+B) = ~A~B (I'm using the ~ for the negation operator, where you use an apostrophe. With that and the rule that ~~A == A you have all you need. Here's the first few lines of the derivation to get to a full conjuctive form.

ABC + ~A~C + ~A~B
= ~~(ABC) + ~~(~A~C) + ~~(~A~B)
= ~(~A + ~B + ~C) + ~~(~A~C) + ~~(~A~B)
= ~(~A + ~B + ~C) + ~(~~A + ~~C) + ~~(~A~B)
= ~(~A + ~B + ~C) + ~(A + C) + ~~(~A~B)
(I'm sure you can do the rest)

An exactly mirror technique gets you to the full disjunctive form.

@Griswolf: Hi i'm also trying to solve this problem but i'm still not gettng it can u explain it to me if possible? Thank You

[griswolf]

I'm lost. What part of ~~(AB) == ~(~A + ~B) did you not get?

[gooradog]

from what u wrote is it possible to get only Or and Not/ And and Not solution? and ~ <- this is same as having like X with Bar on top right?

[griswolf]

Yes, it is possible to convert any boolean expression so it uses only two operators (plus parentheses):AND and NOT (and parens)
OR and NOT (and parens)

Yes, ~B is the same as B under a bar or B'

[gooradog]

~(~A + ~B + ~C) + ~(A + C) + ~~(~A~B)
= ~(~A + ~B + ~C) + ~(A + C) + ~(A + B)
so is this express it as or and not operation?

[griswolf]

= ~(~A + ~B + ~C) + ~(A + C) + ~(A + B) has no AND part: No place where you see AB ... where A or B might each be any expression

[gooradog]

I meant to say if that was "OR and NOT" Operation ^^
But from the "OR and NOT operation" can we convert it to "And and NOT operation?"

[gooradog]

Or do i need to something else for it to be "AND and NOT" Operation?