Boolean Algebra - proof without associativity? I would like to prove the following:
$(x\cdot y) + (\overline{x} + \overline{y}) = 1$
without the Associativity Property. I can't seem to do this algebraically (without truth tables).
 A: $$(x\cdot y) + (\overline x + \overline y) = (x\cdot y)+ (\overline{x\cdot y}) = 1$$
We simply make use of Demorgan's Law, and the identity $p + \overline p = 1$.
A: So I cannot use Associative:
$X + (Y + Z) = (X + Y) + Z = (X + Z) + Y = X + Y + Z $
$(x\cdot y) + (\overline{x} + \overline{y})$
Expand minimized terms using complement 
$(x\cdot y) + (\overline{x} \cdot (\overline{y} + y) + \overline{y} \cdot (\overline{x} + x))$
$(x\cdot y) + (\overline{x}\cdot\overline{y} + \overline{x}\cdot y + \overline{y} \cdot \overline{x} + \overline{y}\cdot x)$
Remove duplicates with Idempotent.
$(x\cdot y) + (\overline{x}\cdot\overline{y} + \overline{x}\cdot y + \overline{y}\cdot x)$
Extract common terms with Redundancy and Complement.
$x \cdot (y + \overline{y}) + \overline{x} \cdot (\overline{y} + y)$
I may of used Associative by removing brackets.  Complement again.
$x + \overline{x}$
$1$
A lot easier using deMorgan's.
Laws and Theorems of Boolean Algebra
A: First note that
\begin{align*}
x + (\overline x + \overline y)
&= 1\cdot (x + (\overline x + \overline y)) \\
&= (x + \overline x)\cdot (x + (\overline x + \overline y)) \\
&= x + (\overline x\cdot (\overline x + \overline y)) \\
&= x + ((\overline x + 0)\cdot (\overline x + \overline y)) \\
&= x + (\overline x + (0\cdot \overline y)) \\
&= x + (\overline x + 0) \\
&= x + \overline x \\
&= 1
\end{align*}
Similarly,
$$ y + (\overline y + \overline x) = 1 $$
So
\begin{align*}
(x\cdot y) + (\overline x + \overline y)
&= (x + (\overline x + \overline y))\cdot (y + (\overline x + \overline y))
\\
&= (x + (\overline x + \overline y))\cdot (y + (\overline y + \overline x))
\\
&= 1\cdot 1 \\
&= 1
\end{align*}
This uses distributivity and the properties of $0$ and $1$, but not associativity and not De Morgan.  (I also used the commutativity of $+$, but it's not essential.)
