Algebraic proof of De Morgan's Theorems Could someone give me an algebraic proof of De Morgan's Theorems?
I already know the graphic proof with the truth table, but I need to understand the algebraic way.

EDIT
I try to explain better. Imagine to have the two theorems:
NOT(a * b) = NOT(a) + NOT(b)
NOT(a + b) = NOT(a) * NOT(b)

What do you answer to someone that tell you: "Show me why this two theorems are verified without using the truth tables"?
 A: Following, e.g. Wikipedia, let us define a boolean algebra to be a set $A$, together with two binary operations $\land$ and $\lor$, a unary operation $'$, and two nullary operations $0$ and $1$, satisfying the following axioms:
$$\begin{align*}
a\lor(b\lor c) &= (a\lor b)\lor c, & a\land(b\land c) &= (a\land b)\land c, &&\text{(associativity)}\\
a\lor b &= b\lor a, & a\land b &= b\land a, &&\text{(commutativity)}\\
a\lor(a\land b) &= a, & a\land (a\lor b) &= a, &&\text{(absorption)}\\
a\lor(b\land c) &= (a\lor b)\land (a\lor c), & a\land(b\lor c) &= (a\land b)\lor (a\land c) &&\text{(distributivity)}\\
a\lor a' &= 1, & a\land a' &= 0. &&\text{(complements)}
\end{align*}$$
You want to use these axioms to prove that $(a\land b)' = a'\lor b'$ and $(a\lor b)' = a'\land b'$.
Lemma 1. $a\land 1 = a$ and $a\lor 0 = a$ for all $a$.
Proof. 
$a \land 1 = a\land(a\lor a') = a$, by complements and absorption; likewise, $a\lor 0 = a\lor(a\land a') = a$ by complements and absorption. $\Box$
Lemma 2. $a\land 0 = 0$ and $a\lor 1 = 1$ for all $a$.
Proof. $a\land 0 = a\land (a\land a') = (a\land a)\land a' = a\land a' = 0$. And $a\lor 1 = a\lor(a\lor a') = (a\lor a)\lor a' = a\lor a' = 1$. $\Box$
Lemma 3. If $a\land b' = 0$ and $a\lor b'=1$, then $a=b$. 
Proof.
$$\begin{align*}
b &= b\land 1\\
&= b\land(a\lor b')\\
&= (b\land a)\lor (b\land b')\\
&= (b\land a)\lor 0\\
&= (b\land a)\lor (a\land b')\\
&= (a\land b)\lor(a\land b')\\
&= a\land (b\lor b')\\
&= a\land 1\\
&= a.\ \Box
\end{align*}$$
Lemma 4. For all $a$, $(a')' = a$.
Proof. By Lemma 3, it suffices to show that $(a')'\land a' = 0$ and $(a')'\lor a' = 1$. But this follows directly by complementation. $\Box$
Theorem. $(a\land b)' = a'\lor b'$.
By Lemmas 3 and 4, 
it suffices to show that $(a\land b)\land (a'\lor b') = 0$ and $(a\land b)\lor (a'\lor b') = 1$; for by Lemma 4, this is the same as proving $(a\land b)\land (a'\lor b')'' =0$ and $(a\land b)\lor (a'\lor b')'' = 1$; by Lemma 3, this gives $(a\land b) = (a'\lor b')'$, and applying Lemma 4 again we get $(a\land b)' = (a'\lor b')'' = a'\lor b'$, which is what we want. 
We have:
$$\begin{align*}
(a\land b)\land(a'\lor b') &= \bigl((a\land b)\land a')\bigr) \lor \bigl((a\land b)\land b') &&\text{(by distributivity)}\\
&= \bigl( (a\land a')\land b\bigr) \lor \bigl( a\land (b\land b')\bigr) &&\text{(associativity and commutativity)}\\
&= ( 0\land b) \lor (a\land 0)\\
&= 0 \lor 0\\
&= 0.
\end{align*}$$
And
$$\begin{align*}
(a\land b)\lor(a'\lor b') &= \bigl( (a\land b)\lor a'\bigr) \lor b'&&\text{(by associativity)}\\
&= \bigl( (a\lor a') \land (b\lor a')\bigr) \lor b'&&\text{(by distributivity)}\\
&= \bigl( 1\land (b\lor a')\bigr) \lor b'&&\text{(by complements)}\\
&= (b\lor a')\lor b'&&\text{(by Lemma 1)}\\
&= (b\lor b')\lor a'&&\text{(by commutativity and associativity)}\\
&= 1\lor a'&&\text{(by complements)}\\
&= 1 &&\text{(by Lemma 2)}.
\end{align*}$$
Since $(a\land b)\land (a'\lor b') = 0$ and $(a\land b)\lor (a'\lor b') = 1$, the conclusion follows. $\Box$
Theorem. $(a\lor b)' = a'\land b'$.
Proof. Left as an exercise for the interested reader. $\Box$
A: The truth tables are a summary of the underlying algebra; without further specification of what you mean by "the algebraical way" I don't know what you would want other than truth tables. You could split it into cases ("if $A=0$ and $B=0$ then $(A+B)' = 0' = 1 = 1 \cdot 1 = A' \cdot B'$", and so on), but this is exactly what the truth table tells you.
A: 
What do you answer to someone who says: "Show me why this two theorems are verified without using the truth tables."?

The simplest possible example that I can think of for the de Morgan laws is:
If you ask:
 "When does an $AND$ (or $\land$) statement is not $true$?", then the answer is: "If either of the two statements involved is $false$."
The above is described symbolically with: $$\lnot(a \land b) = \lnot a \lor \lnot b $$
Similarly, if you ask:
"When does an $OR$ (or $\lor$) statement is not $true$?", then the answer is: "When both of the involved statements are $false$."
Which symbolically is:
$$\lnot(a \lor b) = \lnot a \land \lnot b$$
Thus, the de Morgan laws could be thought of as a criteria for the result of the logical operators $OR$ and $AND$ to be $false$. 

More convoluted answer:
Considering: 

The statements $\lnot \forall x \in U(p(x))$ and $\exists x \in U(\lnot p(x))$ are equivalent.

The de Morgan laws could be thought of as a reduction of the relationship that negation, $\neg$, gives between "for all", $\forall$, and "there exists", $\exists$, statements, from a potentially infinite many statements about a infinite universe to finite number of statements.
A: Transferring the problem from the Boolean algebra $(\mathbb Z_2,\neg,\vee,\wedge)$ to the Boolean ring $(\mathbb Z_2,1,+,\cdot)$, where $1$ means TRUE, $+$ means EXCLUSSIVE OR and $\cdot$ means AND and using the equivalences
$\neg a \iff (1+a)$
$a\vee b \iff (a+b+ab)$
$a\wedge b \iff (a\cdot b)$
makes the task very algebraic and rather straightforward.
$\neg(a\wedge b)\iff (1+ab)$
$(\neg a \vee \neg b)\iff ((1+a)+(1+b)+(1+a)(1+b))=a+b+1+a+b+ab=1+ab$
(remember that $x+x=0$)
and
$\neg(a\vee b)\iff (1+(a+b+ab))$
$(\neg a)\wedge\neg b)(1+a)(1+b)=1+a+b+ab$
