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 $\displaystyle (B(B+C))'$ 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...
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?
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.
* = 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)