Using Demorgan's Theoroem to Simplify C'+D'+A'B'CD

[jisbon]

Homework Statement

Simplify Boolean expression C'+D'+A'B'CD

Relevant Equations

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Hi all,
I'm new to Demorgan's theorem and hence am dumbfounded on how people actually simplify the equation by just looking at it. I was wondering if there are any tricks to doing so and I will appreciate it if someone can teach me.

In the example of this question, what I can only simply come out with is C'+D'+A'B'CD = (CD)'+ A'B'CD. I can't seem to get around my head what to do next. I do know the following rules though:
A + A = A
A' + A = 1
(A+BC)=(A+B)(A+C)
(A+B)' = A'B'
and
A'+B'= (AB)'

Any tips/tricks for this? Thank you so much!
' = inverse fyi.

 

[haruspex]

One problem with such questions is agreeing what constitutes a simpler form.

The only trick I know is to decipher its meaning. In the present case, C'+D'+A'B'CD
is true if not C, or not D, or (C and D and ...)
So it is true whenever not C, and the only other cases we have to consider are where C is true, so any occurrence of C in an AND beyond that is redundant:
C'+D'+A'B'D
Likewise, D:
C'+D'+A'B'

To get this from what you had and the rules, use (a+bc)=(a+b)(a+c), substituting CD for a, A'B' for b and (CD)' for c.

 

[sysprog]

The character played by John Cleese on 'Fawlty Towers' said his to wife "they ought to put you on Mastermind -- special category: the bleedin' obvious" . . .

I think that DeMorgan's rule for by negation interchanging disjunction and conjunction is so obvious that it should not have been named after anyone.

@jisbon: your use of arithmetical symbols and apostrophes instead of logic symbols is non-standard -- please go to the $\LaTeX$ Guide page here or do a search engine lookup $\cdots$ I'm not a Mentor/Moderator here, but I think that you have shown enough work for the other members to be allowed to be of assistance, and I'll be glad to be of assistance if you even merely use $\land$ instead of using + and please find something else, such as brackets or braces, that means whatever you mean by the apostrophes -- sorry to be such a stickler about syntax -- I guess that I'm a little like a compiler in that regard $\cdots$