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).

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.)
