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.
$$ \begin{array}{ |c|c| } \hline {\text{Axioms}}\\ \hline \text{Property of 0} & X + 0 = X \space ;\space X\cdot0 = 0 \\ \text{Property of 1} & X + 1 = 1 \space;\space X\cdot1 = X\\ \text{Idempotence Law} & X + X = X \space;\space X\cdot X = X\\ \text{Involution Law} & \overline{\overline X} = X\\ \text{Complementarity Law} & X + \overline X = 1 \space;\space X\cdot\overline X = 0\\ \text{Commutative Law} & X+Y = Y+X \space;\space X\cdot Y = Y\cdot X\\ \text{Associative Law} & (X+Y)+Z = X+(Y+Z) \space;\space (X\cdot Y)\cdot Z = X\cdot (Y\cdot Z)\\ \text{Distributive Law} & X(Y+Z) = XY + XZ \space;\space X + YZ = (X+Y)(X+Z)\\ \text{Absorption Law} & X + XY = X \space;\space X(X+Y) = X\\ \text{Other (3rd Distributive)} & X + \overline XY = X+Y\\ \hline \end{array} $$
$\endgroup$ 62 Answers
$\begingroup$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.
$\endgroup$ 5 $\begingroup$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
