Boolean Algebra Proof for a + a = a and (a * b)' = a' + b'

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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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3 Answers

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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}

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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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from right side

=a

=a+0

=a+(a.a')

=(a+a).(a+a')

=(a+a).1

=a+a

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