simplifying following function

Question

F(A, B, C, D) = A'B'C' + AC' + ACD + ACD' + A'B'D'

I first did AC(D+D') which got me just AC, and then A(C+C') which left me with just A. The function is now A'B'C'+ A + A'B'D'. I am unsure if I can do things like pulling the A out of A'B'D' to do A+A. If it isn't, I am stuck, what more can I do?

Note that $X + X'Y = X + Y$. – Fabio Somenzi Feb 21 '18 at 02:01 — Fabio Somenzi, Feb 21 '18 at 02:01
oh I see, so A+A'B'D' would be A+B'D' @FabioSomenzi – Kytex Feb 21 '18 at 02:09 — Kytex, Feb 21 '18 at 02:09

score 0 · Accepted Answer · answered Feb 21 '18 at 02:30

0

$$A+A'B'D' \overset{Distribution}{=}$$

$$(A+A')(A+B')(A+D') \overset{Complement}{=}$$

$$1(A+B')(A+D') \overset{Identity}{=}$$

$$(A+B')(A+D') \overset{Distribution}{=}$$

$$A+B'D'$$

answered Feb 21 '18 at 02:30

Bram28

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simplifying following function

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