Verify Demorgan's Law Algebraically If $\overline X \equiv \text { not }X$,
De Morgan's Laws are stated as:


*

*$ \overline{(A + B)}=  \overline A\cdot \overline B$ 

*$ \overline{(A\cdot B)} = \overline A + \overline B$
Verify the above laws algebraically.
I can prove this using truth tables and logic gates but algebraically, I don't know any intuitive way to prove it.
$$
\begin{array}{ |c|c| }
  \hline
   {\text{Axioms}}\\
  \hline
   \text{Property of 0} & X + 0 = X \space ;\space X\cdot0 = 0 \\
   \text{Property of 1} & X + 1 = 1 \space;\space X\cdot1 = X\\
   \text{Idempotence Law} & X + X = X \space;\space X\cdot X = X\\
   \text{Involution Law} & \overline{\overline X} = X\\
   \text{Complementarity Law} & X + \overline X = 1 \space;\space X\cdot\overline X = 0\\
   \text{Commutative Law} & X+Y = Y+X \space;\space X\cdot Y = Y\cdot X\\
   \text{Associative Law} & (X+Y)+Z = X+(Y+Z) \space;\space (X\cdot Y)\cdot Z = X\cdot (Y\cdot Z)\\
   \text{Distributive Law} & X(Y+Z) = XY + XZ \space;\space X + YZ = (X+Y)(X+Z)\\
   \text{Absorption Law} & X + XY = X \space;\space X(X+Y) = X\\
   \text{Other (3rd Distributive)} & X + \overline XY = X+Y\\
  \hline
\end{array}
$$
 A: By Complementarity Law, 
$$P + \overline P = 1 \space\text{ and }\space P \cdot \overline P = 0$$
(Note: I shall only be using $P + \overline P = 1$ as its dual is automatically true)

First Law::
DeMorgan's $1^{\text{st}}$ law states $\overline{X+Y} = \overline X \cdot \overline Y$

It is sufficient to prove that $(X + Y) +  \overline X \cdot \overline Y = 1$
$$
\begin{align}
\text{LHS} &= Y + (X + \overline X \cdot \overline Y)\\ 
&= Y +  X + \overline Y\\
&= (Y+\overline Y) + X\\
&= 1 + X \\&= 1 = \text{RHS}
\end{align}
$$

Second Law:: DeMorgan's $2^{\text{nd}}$ Law states that $\overline{X\cdot Y} = \overline X + \overline Y$

It is sufficient to prove that $X\cdot Y + (\overline X + \overline Y) = 1$
$$
\begin{align}
\text{LHS} &= \overline Y + (\overline X + \overline{\overline X}\cdot Y)\\
&= \overline Y + (\overline X + Y)\\
&= (Y + \overline Y) + \overline X\\
&= 1 + \overline X\\
&= 1 = \text{RHS}
\end{align}
$$

Hence, DeMorgan's Laws are verified algebraically.
A: Lemma: $ A + B = 1 \land AB = 0 \implies A = \overline B $
Proof:
Given $ A + B = 1 $, then:
$ (A + B) \overline B = 1 \cdot \overline B = \overline B $
$\implies A \overline B + B \overline B = A \overline B + 0 = A \overline B = \overline B $   (1)
Given $ AB = 0 $, we combine it with (1) to obtain:
$ A \overline B + AB = \overline B + 0 $
$\implies A (\overline B + B) = A \cdot 1 = A = \overline B$   QED
 
First law:
$ (X + Y) + \overline X \overline Y = Y + (X + \overline X \overline Y) = $
  $ = Y + (X + \overline Y) = (Y + \overline Y) + X = $
  $ = 1 + X = 1 $   (2)
$ (X + Y) \cdot \overline X \overline Y = X \cdot \overline X \overline Y + Y \cdot \overline X \overline Y = $
  $ = (X \overline X) \cdot \overline Y + (Y \overline Y) \cdot \overline X = 0 \cdot \overline Y + 0 \cdot \overline X = $
  $ = 0 + 0 = 0 $   (3)
By (2), (3), and the lemma, we have $ \overline { X + Y } = \overline X \overline Y $ -- QED
 
Second law:
$ \overline { X Y } = \overline { { \overline { \overline X } } \cdot { \overline { \overline Y } } } = $   (by First law)
  $ = \overline { \overline { \overline X + \overline Y } } = \overline X + \overline Y $   QED
 
Notes


*

*Contrary to the approach in the first answer, it is necessary to prove both that $ A + B = 1 $ and $ AB = 0 $ in order to claim that $ A = \overline B $.

*Absorption Law and 3rd Distributive are redundant as axioms, following from Property of 1, Complementarity Law, Idempotence Law, and Associative Law.

*Idempotence Law is redundant as an axiom, following from Property of 1 and Distributive Law
