Thread: Boolean Reduction Question

1. Boolean 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

2. Originally Posted by anderson
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
No, I think that your converting $(B(B+C))'$ to $B+B'C'$ is wrong already. Instead, I get the following

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

This is because we have that $B\leq B+C$ and because from $U\leq V$ we can always conclude that $UV=U$ (and that $U+V=V$).
In our case, this means that from $B\leq B+C$ we may conclude that $B(B+C)=B$.

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

(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 $B'$ in your notation (or $\overline{B}$ in my notation) - but do the rest of the calculation yourself...

3. pls 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

4. 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?

5. HI 0113

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

Tq & rgds

6. Originally Posted by O113
So! You get:
((A' + B') B') = A'B' + B'B' = A'B' + B' = B' (de Morgan again)

Did I get it right?
Well, yes, but you could have got that result more easily, in a single step, like this: $(A' + B') B'=B'$

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

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

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

7. * = 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)