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Is a complement + 1 = 1? For example A' + 1 = 0;

I was thinking it was (I'm new to boolean algebra) since A' = 0, and 0 + 1 in boolean algebra is just 1. Of course, A can be anything, but assuming this is a single variable like B being represented as A, compared to ABCD being represented as A.

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What do the axioms say is the value of $x+1$ for any $x$? – Dilip Sarwate Apr 4 '12 at 3:35
Dilip, x + 1 = 1 for any x. But doesn't the x have to be non-complement? Or can x even be complements? – ShrimpCrackers Apr 4 '12 at 3:37
No, that is for any $x$ in the algebra. – Brian M. Scott Apr 4 '12 at 3:39
Sorry, Brian not sure I follow. "No" to what? – ShrimpCrackers Apr 4 '12 at 3:41
Please re-read Demorgan's Laws very carefully. Does one of them really say $X^\prime Y^\prime = X^\prime + Y^\prime$ as you contend, or does it say $(XY)^\prime = X^\prime + Y^\prime$ ?? – Dilip Sarwate Apr 4 '12 at 4:00

Answered in the comments: yes, (something) + 1 = 1 in a Boolean algebra. It does not matter if (something) was obtained as a complement of an element, or in another way. Dilip Sarwate made a good suggestion:

If it makes you feel better, begin your proof with the statement: "Let $x$ denote $A'$. Then, since $x+1=1$ by Axiom $\dots$, we have that $A'+1=1$."

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