I'm just getting into digital design, and I've been hit with a pretty hard logical expression.
$ [A (B+C)' + BDE][A' + CE]$
I don't know if it's required, and more importantly to what extent it is possible to simplify this, but I had a go at it and used the distributive property to see where that would get me.
$ AA'(B+C)' + ACE(B+C)' + BDEA' + BDEEC$
A and A' contradict, zeroing out that first term.
$ 0 + ACE(B+C)' + BDEA' + BDEC$
I see I can factor out BDE, if that helps.
$ACE(B+C)' + BDE(A' + C)$
But even then, I don't know how much more I can do. Could someone please see what I might not?
Many thanks!
-Jon
First off, 'simplify' is a little ambiguous: is a statement more simplified if it is in DNF or CNF, or is it more simplified when it has the least number of symbols (but not necessarily in DNF or CNF)?
Anyway:
$$0 + ACE(B+C)' + BDEA' + BDEC = \text{ (Identity)}$$
$$ACE(B+C)' + BDEA' + BDEC = \text{ (DeMorgan)}$$
$$ACEB'C' + BDEA' + BDEC = \text{ (Complement)}$$
$$0 + BDEA' + BDEC = \text{ (Identity)}$$
$$BDEA' + BDEC$$
which is as 'simple' as you can get it in DNF. But you can indeed take out the $BDE$:
$$BDE(A' + C)$$
and now it is in CNF (and with fewer symbols!)