Prove, for any element $a$ in a boolean algebra expression, that $a + a = a$. Prove also, for any two elements, $a$ and $b$, of a boolean algebra expression, that $(a * b)' = a' + b'$.
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Whats $+$? Etc, juse mathjax. – Zelos Malum Jan 24 '17 at 07:03
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I suppose $+$ stands for $\lor$, $\cdot$ for $\land$ – b00n heT Jan 24 '17 at 07:44
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I proved it. That's all you wanted, right? (Please ask, instead of commanding, and use proper formatting.) – The Count Feb 03 '17 at 03:02
3 Answers
Idempotent law a + a = a
Proof: x + x
= (x + x) • 1
= (x + x) • (x + x')
= x + (x • x')
= x + 0 = x
And for other prove see de-morgan's law.
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I can't prove DeMorgan's Law (the second equation you gave) because, quite frankly, I don't know how, but I can prove your first expression. I will be using symbols from boolean algebra, not formal logic.
In boolean algebra addition represents or ($\lor$), multiplication represents and ($\land$), and a bar ($\bar{}$) over an element represents a negation ($\neg$).
Proof of $x + x = x$
\begin{align} x + x &= x && \text{Original Equation}\\ x + x &= x + 0 && \text{Identity Law}\\ x + x &= x + x\bar{x} && \text{Zero Property}\\ x + x &= (x + x)(x + \bar{x}) && \text{Distributive Law}\\ x + x &= (x + x) * 1 && \text{Unit Property}\\ x + x &= x + x && \text{Identity Law}\\ \end{align}
Laws and Properties Used
Identity Law:
$$ x + 0 = x $$ $$ x * 1 = x $$
Distributive Law
$$ x + yz = (x + y)(x + z) $$ $$ x(y + z) = xy + xz $$
Zero Property
$$ x\bar{x} = 0 $$
Unit Property
$$ x + \bar{x} = 1 $$
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I'll ask one of my discrete mathematics professors soon how to prove DeMorgan's Law. If I'm able to understand it myself, I will edit this answer and add the proof and any laws/properties used in addition to the ones that I used for the first proof. – George Edward Shaw IV Oct 31 '19 at 09:47