Member Avatar for 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

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Member Avatar for 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.

Member Avatar for 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

Member Avatar for 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

Member Avatar for griswolf

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

Edited by griswolf because: n/a
Member Avatar for 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?

Member Avatar for 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'

Member Avatar for 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?

Member Avatar for 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

Member Avatar for 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?"

Member Avatar for gooradog

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

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