<s>[INST] 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).
 [/INST] 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
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