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 $\displaystyle (B(B+C))'$ to $\displaystyle B+B'C'$ is wrong already. Instead, I get the following

$\displaystyle (B(B+C))'=B'$

This is because we have that $\displaystyle B\leq B+C$ and because from $\displaystyle U\leq V$ we can always conclude that $\displaystyle UV=U$ (and that $\displaystyle U+V=V$).

In our case, this means that from $\displaystyle B\leq B+C$ we may conclude that $\displaystyle B(B+C)=B$.

Similarly I believe that your converting $\displaystyle (AB)'$ to $\displaystyle A'B'$ is wrong. By De Morgan's Laws you get, instead, that $\displaystyle (AB)'=A'+B'$.

(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 $\displaystyle B'$ in your notation (or $\displaystyle \overline{B}$ 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: $\displaystyle (A' + B') B'=B'$

Why? - Let me explain this by converting the whole thing to set theory: $\displaystyle (\overline{A}\cup\overline{B})\cap \overline{B}=\overline{B}$

Which is to say: if you intersect the superset $\displaystyle \overline{A}\cup\overline{B}$ with its subset $\displaystyle \overline{B}$ you get ... just the subset. Or, to use new symbols: from $\displaystyle Y\subseteq X$ you can conclude that $\displaystyle X\cap Y= Y$ (and that $\displaystyle X\cup Y=X$). In other words, $\displaystyle \cap$ behaves like a greatest lower-bound and $\displaystyle \cup$ behaves like a least upper-bound.

Or to explain it in propositional calculus: $\displaystyle (\neg A\vee \neg B)\wedge \neg B\Leftrightarrow \neg B$, because this follows from $\displaystyle \neg A\vee \neg B\Rightarrow \neg B$. Again: $\displaystyle \wedge$ behaves like a greatest lower-bound and $\displaystyle \vee$ behaves like a least upper-bound.

I don't know about the symbolism the OP uses. Maybe one should write that $\displaystyle (A'+B')B'=B'$ holds, because $\displaystyle B'\leq A'+B'$? - But I just don't know how the relationship $\displaystyle \subseteq$ for sets and $\displaystyle \Rightarrow$ 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)