Hi everyone

Question:

(AB)'(B(B+C))'

My working:

=A'B'(B'B'+B'C')

=(A'B')(B+B'C')

=(A'B')(B'C')

=B'(A'+C')

did i get it right....can someone help me please???

tq & rgds

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- Aug 4th 2009, 03:36 AMandersonBoolean Reduction Question
Hi everyone

Question:

**(AB)'(B(B+C))'**

**My working:**

=A'B'(B'B'+B'C')

=(A'B')(B+B'C')

=(A'B')(B'C')

=B'(A'+C')

did i get it right....can someone help me please???

tq & rgds - Aug 4th 2009, 08:46 AMFailure
No, I think that your converting to is wrong already. Instead, I get the following

This is because we have that and because from we can always conclude that (and that ).

In our case, this means that from we may conclude that .

Similarly I believe that your converting to is wrong. By De Morgan's Laws you get, instead, that .

(Actually, I feel slightly uneasy about your notation. I'm more the "logical chap" comming from the propositional calculus.)

If I'm not mistaken, the given expression is equivalent to in your notation (or in my notation) - but do the rest of the calculation yourself... - Aug 4th 2009, 10:45 PMandersonpls help me...
HI everyone

Question:

**(A.B)'.(B(B+C)'**=

(AB)' = A'+B',B(B+C)=(BC)'

A' + B' + (BC)'

=A'+B'+B'+C'

=A'+C'

Pls help advise.

Tq & rgds - Aug 5th 2009, 03:10 AMO113
This looks alot like digital technology rather than calculus-logic. Tbf, they are the same thing, just that they use a slightly different form of notation. Anyhow! I'm going to give this a try, but I wouldn't bet my life on the solution:

First, I'd separate it into 2 parts, effectively:

(A * B)'

and

(B (B + C))'

(A * B)' = (A' + B')

(B (B + C))' = (B' + (B' * C' ) = (This can be reduced to B' per de Morgan laws, they all have fancier names but I can never remember them)

So! You get:

((A' + B') B') = A'B' + B'B' = A'B' + B' = B' (de Morgan again)

Did I get it right? - Aug 5th 2009, 06:50 AManderson
HI 0113

Your answer looks very accurate,thank you. Tq for all the help & support..any other ideas,please share...

Tq & rgds - Aug 5th 2009, 08:34 AMFailure
Well, yes, but you could have got that result more easily, in a single step, like this:

Why? - Let me explain this by converting the whole thing to set theory:

Which is to say: if you intersect the superset with its subset you get ... just the subset. Or, to use new symbols: from you can conclude that (and that ). In other words, behaves like a greatest lower-bound and behaves like a least upper-bound.

Or to explain it in propositional calculus: , because this follows from . Again: behaves like a greatest lower-bound and behaves like a least upper-bound.

I don't know about the symbolism the OP uses. Maybe one should write that holds, because ? - But I just don't know how the relationship for sets and for propositions is treated in his formalism. - Aug 5th 2009, 04:01 PMO113
* = AND

A ring with a cross in it (can't find the latex command) = Exclusive Or

+ = OR

' = Inverse

The rest you kinda skip since they aren't needed for basic circuit technology. You rewrite logical expressions by using a = instead of caring about =>, ->, <=>

Example:

(A + B) means A or B

A * B means A and B

A' means "Not A"

The rest you can figure out for yourself xD

I can't say that I really know enough of this to actually give you a good explanation about how the arrows are used for more difficult problems. I've always assumed that it's the same as in calculus (loosely based on a short notation in an introductory course)