# Boolean Simplification (ABCD)' + ((CD)'(B+D)'

I have to simplify (ABCD)' + ((CD)'(B+D)' function using boolean algebra. I simplified it using a truth table and got A'+B' + CD. There is a law I am having trouble understanding

Apparently (AB)' + (B + D)' Simplifies to A'+ B'

Can anyone help me with these 2 instructions? Thank You.

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I'm not sure what our OP Tokenizer means by "these two instructions", unless the phrase refers to the two expressions $(AB)' + (B + D)'$ and $A' + B'$. But it is certainly true that

$(AB)' + (B + D)' = A' + B', \tag{1}$

for we have, by de Morgan and the identity $1 + D = 1$,

$(AB)' + (B + D)' = A' + B' + B'D' = A' + B'(1 + D') = A' + B'(1) = A' + B'. \tag{2}$

QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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Let There Be Light? haha. Thanks for the help. –  The Tokenizer Feb 1 '14 at 3:51
@ Tokenizer: glad to. Hey, Fiat Lux!!! is a good motto for switching theory, no? All those series/parallel combos of toggles . . . with a bulb at the end! ;-)! –  Robert Lewis Feb 1 '14 at 3:53
(AB)' = (A'+B')   (demorgan's law)
(B+D)' = (B'D')   (demorgan's law)
B'+B'D' = B'


so

(AB)' + (B+D)' = A'+B'+B'D' = A'+B'

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You've made it very clear for me, Thank You. –  The Tokenizer Feb 1 '14 at 3:53