I started with the definition of boolean algebras by Huntington, which is:
A boolean algebra is a set $B$ with two operations on $B$, so that for all $a \in B$, $b \in B$ and $c \in B$ holds:
Commutativity: \begin{equation}\label{1.1.1}\tag{1.1.1} a \land b = b \land a \end{equation} \begin{equation}\label{1.1.2}\tag{1.1.2} a \lor b = b \lor a \end{equation} Distributivity: \begin{equation}\label{1.1.3}\tag{1.1.3} a \land \left(b \lor c \right) = \left(a \land b\right) \lor \left(a \land c \right) \end{equation} \begin{equation}\label{1.1.4}\tag{1.1.4} a \lor \left(b \land c \right) = \left(a \lor b\right) \land \left(a \lor c \right) \end{equation} Existence of neutral elements: There are elements $0 \in B$ and $1 \in B$, so that: \begin{equation}\label{1.1.5}\tag{1.1.5} a \land 1 = a \end{equation} \begin{equation}\label{1.1.6}\tag{1.1.6} a \lor 0 = a \end{equation} Existence of complements: For every $a\in B$ there is $\neg a\in B$, so that: \begin{equation}\label{1.1.7}\tag{1.1.7} a \land \neg a = 0 \end{equation} \begin{equation}\label{1.1.8}\tag{1.1.8} a \lor \neg a = 1 \end{equation}
I wanted to prove, that from this definition everything stated here follows. I was able to do this, but I used the associativity to finally prove de morgan's law. Nevertheless I am pretty sure, that you don't need associativity to prove de morgan's law. Could one give me a hint on how to achieve that?
For everbody interested in how I've proven everything (de morgan's law is corollary $\ref{2.11}$):
Corollary 2.1 $\label{2.1}\tag{2.1}$
There is one and only one $0 \in B$.
Be $0_1 \in B$ and $0_2 \in B$ zero elements. It holds: \begin{equation*} 0_1 \overset{\ref{1.1.6}}{=} 0_1 \lor 0_2 \overset{\ref{1.1.2}}{=} 0_2 \lor 0_1 \overset{\ref{1.1.6}}{=} 0_2 \end{equation*} $\Box$
Corollary 2.2 $\label{2.2}\tag{2.2}$
There is one and only one $1 \in B$.
Be $1_1 \in B$ and $1_2 \in B$ one elements. It holds: \begin{equation*} 1_1 \overset{\ref{1.1.5}}{=} 1_1 \land 1_2 \overset{\ref{1.1.1}}{=} 1_2 \land 1_1 \overset{\ref{1.1.5}}{=} 1_2 \end{equation*} $\Box$
Corollary 2.3 $\label{2.3}\tag{2.3}$
There is one and only one $\neg a \in B$ for each $a \in B$.
Be $a \in B$ and be $a_1 \in B$ and $a_2 \in B$ complements of $a$. It holds: \begin{equation*} a_1 \overset{\ref{1.1.6}}{=} a_1 \lor 0 \overset{\ref{1.1.7}}{=} a_1 \lor \left(a \land a_2 \right) \overset{\ref{1.1.4}}{=} \left(a_1 \lor a \right) \land \left(a_1 \lor a_2 \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.2}}{=} \left(a \lor a_1 \right) \land \left(a_1 \lor a_2 \right) \overset{\ref{1.1.8}}{=} 1 \land \left(a_1 \lor a_2 \right) \overset{\ref{1.1.8}}{=} \left(a \lor a_2 \right) \land \left(a_1 \lor a_2 \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.2}}{=} \left(a_2 \lor a \right) \land \left(a_2 \lor a_1 \right) \overset{\ref{1.1.4}}{=} a_2 \lor \left(a \land a_1 \right) \overset{\ref{1.1.7}}{=} a_2 \lor 0 \overset{\ref{1.1.6}}{=} a_2 \end{equation*} $\Box$
Corollary 2.4 $\label{2.4}\tag{2.4}$
For $a \in B$ holds: $$\neg \neg a = a$$
Be $a \in B$, be $\neg a \in B$ the complement of $a$ and be $\neg \neg a \in B$ the complement of $\neg a$. It holds: \begin{equation*} \neg \neg a \overset{\ref{1.1.6}}{=} \neg \neg a \lor 0 \overset{\ref{1.1.7}}{=} \neg \neg a \lor \left(a \land \neg a \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.4}}{=} \left(\neg \neg a \lor a \right) \land \left(\neg \neg a \lor \neg a \right) \overset{\ref{1.1.2}}{=} \left(a \lor \neg \neg a \right) \land \left(\neg a \lor \neg \neg a \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.8}}{=} \left(a \lor \neg \neg a \right) \land 1 \overset{\ref{1.1.8}}{=} \left(a \lor \neg \neg a \right) \land \left(a \lor \neg a \right) \overset{\ref{1.1.4}}{=} a \lor \left(\neg \neg a \land \neg a \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.1}}{=} a \lor \left(\neg a \land \neg \neg a \right) \overset{\ref{1.1.7}}{=} a \lor 0 \overset{\ref{1.1.6}}{=} a \end{equation*} $\Box$
Corollary 2.5 $\label{2.5}\tag{2.5}$
It holds $$\neg 0 = 1$$ and $$\neg 1 = 0$$
\begin{equation*} \neg 0 \overset{\ref{1.1.6}}{=} \neg 0 \lor 0 \overset{\ref{1.1.5}}{=} \neg 0 \lor \left(0 \land 1 \right) \overset{\ref{1.1.4}}{=} \left(\neg 0 \lor 0 \right) \land \left(\neg 0 \lor 1 \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.2}}{=} \left(0 \lor \neg 0 \right) \land \left(1 \lor \neg 0 \right) \overset{\ref{1.1.8}}{=} 1 \land \left(1 \lor \neg 0 \right) \overset{\ref{1.1.6}}{=} \left(1 \lor 0 \right) \land \left(1 \lor \neg 0 \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.4}}{=} 1 \lor \left(0 \land \neg 0 \right) \overset{\ref{1.1.7}}{=} 1 \lor 0 \overset{\ref{1.1.6}}{=} 1 \end{equation*} With \ref{2.4} we have: \begin{equation*} 0 = \neg \neg 0 = \neg 1 \end{equation*} $\Box$
Corollary 2.6 $\label{2.6}\tag{2.6}$
For every $a \in B$ holds: $$a \lor a = a$$ and $$a \land a = a$$
Be $a \in B$. It holds: \begin{equation*} a \lor a \overset{\ref{1.1.5}}{=} \left(a \lor a \right) \land 1 \overset{\ref{1.1.8}}{=} \left(a \lor a \right) \land \left(a \lor \neg a \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.4}}{=} a \lor \left(a \land \neg a \right) \overset{\ref{1.1.7}}{=} a \lor 0 \overset{\ref{1.1.6}}{=} a \end{equation*} By the duality principle follows the second part. $\Box$
Corollary 2.7 $\label{2.7}\tag{2.7}$
For all $a \in B$ holds: $$a \lor 1 = 1$$ and $$a \land 0 = 0$$
Be $a \in B$. It holds: \begin{equation*} a \lor 1 \overset{\ref{1.1.5}}{=} \left(a \lor 1 \right) \land 1 \overset{\ref{1.1.1}}{=} 1 \land \left(a \lor 1 \right) \overset{\ref{1.1.8}}{=} \left(a \lor \neg a \right) \land \left(a \lor 1 \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.4}}{=} a \lor \left(\neg a \land 1 \right) \overset{\ref{1.1.5}}{=} a \lor \neg a \overset{\ref{1.1.8}}{=} 1 \end{equation*} By the duality principle follows the second part. $\Box$
Corollary 2.8 $\label{2.8}\tag{2.8}$
For all $a \in B$ and $b \in B$ holds: $$a \land \left(a \lor b \right) = a$$ and $$a \lor \left(a \land b \right) = a$$
Be $a \in B$ and $b \in B$. It holds: \begin{equation*} a \land \left(a \lor b \right) \overset{\ref{1.1.6}}{=} \left(a \lor 0 \right) \land \left(a \lor b \right) \overset{\ref{1.1.4}}{=} a \lor \left(0 \land b \right) \overset{\ref{1.1.1}}{=} a \lor \left(b \land 0 \right) \end{equation*} \begin{equation*} \overset{\ref{2.7}}{=} a \lor 0 \overset{\ref{1.1.6}}{=} a \end{equation*} By the duality principle follows the second part. $\Box$
Lemma 2.9 $\label{2.9}\tag{2.9}$
Be $a \in B$, $b \in B$ and $c \in B$ with $$a \lor c = b \lor c$$ and $$a \lor \neg c = b \lor \neg c$$ It follows: $$a = b$$ Also holds, that from $$a \land c = b \land c$$ and $$a \land \neg c = b \land \neg c$$ follows: $$a = b$$
Be $a \in B$, $b \in B$ and $c \in B$ with $$a \lor c = b \lor c$$ and $$a \lor \neg c = b \lor \neg c$$ It holds: \begin{equation*} a \overset{\ref{1.1.6}}{=} a \lor 0 \overset{\ref{1.1.7}}{=} a \lor \left(c \land \neg c \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.4}}{=} \left(a \lor c \right) \land \left(a \lor \neg c \right) = \left(b \lor c \right) \land \left(b \lor \neg c \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.4}}{=} b \lor \left(c \land \neg c \right) \overset{\ref{1.1.7}}{=} b \lor 0 \overset{\ref{1.1.6}}{=} b \end{equation*} By the duality principle follows the second part. $\Box$
Corollary 2.10 $\label{2.10}\tag{2.10}$
For all $a \in B$, $b \in B$ and $c \in B$ holds: $$a \land \left(b \land c \right) = \left(a \land b\right) \land c$$ and $$a \lor \left(b \lor c \right) = \left(a \lor b\right) \lor c$$
Be $a \in B$, $b \in B$ and $c \in B$. Be also $$x := a \land \left(b \land c \right)$$ and $$y := \left(a \land b\right) \land c$$ It holds: \begin{equation*} a \lor x = a \lor \left(a \land \left(b \land c \right) \right) \overset{\ref{1.1.4}}{=} \left(a \lor a \right) \land \left(a \lor \left(b \land c \right) \right) \end{equation*} \begin{equation*} \overset{\ref{2.6}}{=} a \land \left(a \lor \left(b \land c \right) \right) \overset{\ref{2.8}}{=} a \end{equation*} Also holds: \begin{equation*} a \lor y = a \lor \left(\left(a \land b\right) \land c \right) \overset{\ref{1.1.4}}{=} \left(a \lor \left(a \land b\right) \right) \land \left(a \lor c \right) \end{equation*} \begin{equation*} \overset{\ref{2.8}}{=} a \land \left(a \lor c \right) \overset{\ref{2.8}}{=} a \end{equation*} All in all: $$a \lor x = a \lor y$$ Also holds: \begin{equation*} \neg a \lor x = \neg a \lor \left(a \land \left(b \land c \right) \right) \overset{\ref{1.1.4}}{=} \left(\neg a \lor a \right) \land \left(\neg a \lor \left(b \land c \right) \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.2}, \ref{1.1.8}, \ref{1.1.1}, \ref{1.1.5}}{=} \neg a \lor \left(b \land c \right) \end{equation*} Also holds: \begin{equation*} \neg a \lor y = \neg a \lor \left(\left(a \land b\right) \land c \right) \overset{\ref{1.1.4}}{=} \left(\neg a \lor \left(a \land b\right) \right) \land \left(\neg a \lor c \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.4}}{=} \left(\left(\neg a \lor a \right) \land \left(\neg a \lor b \right) \right) \land \left(\neg a \lor c \right) \overset{\ref{1.1.2}, \ref{1.1.8}}{=} \left(1 \land \left(\neg a \lor b \right) \right) \land \left(\neg a \lor c \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.1}, \ref{1.1.5}}{=} \left(\neg a \lor b \right) \land \left(\neg a \lor c \right) \overset{\ref{1.1.4}}{=} \neg a \lor \left(b \land c \right) \end{equation*} All in all: $$\neg a \lor x = \neg a \lor y$$ With \ref{2.9} we get $x = y$, what we wanted to prove. By the duality principle follows the second part. $\Box$
Corollary 2.11 $\label{2.11}\tag{2.11}$
For all $a \in B$ und $b \in B$ holds: $$\neg \left(a \land b \right) = \neg a \lor \neg b$$ and $$\neg \left(a \lor b \right) = \neg a \land \neg b$$
Be $a \in B$ and $b \in B$. It holds: \begin{equation*} \left(a \land b \right) \land \left(\neg a \lor \neg b \right) \overset{\ref{2.10}}{=} a \land \left( b \land \left(\neg a \lor \neg b \right) \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.3}}{=} a \land \left(\left(b \land \neg a \right) \lor \left(b \land \neg b \right) \right) \overset{\ref{1.1.7}}{=} a \land \left(\left(b \land \neg a \right) \lor 0 \right) \overset{\ref{1.1.6}}{=} a \land \left(b \land \neg a \right) \end{equation*} \begin{equation*} \overset{\ref{1.1.1}, \ref{2.10}}{=} \left( a \land \neg a \right) \land b \overset{\ref{1.1.7}}{=} 0 \land b \overset{\ref{1.1.1}, \ref{2.7}}{=} 0 \end{equation*} Also holds: \begin{equation*} \left(a \land b \right) \lor \left(\neg a \lor \neg b \right) \overset{\ref{2.10}}{=} \left( \left(a \land b \right) \lor \neg a \right) \lor \neg b \end{equation*} \begin{equation*} \overset{\ref{1.1.2}, \ref{1.1.4}}{=} \left( \left(\neg a \lor a \right) \land \left(\neg a \lor b \right) \right) \lor \neg b \end{equation*} \begin{equation*} \overset{\ref{1.1.2}, \ref{1.1.8}}{=} \left( 1 \land \left(\neg a \lor b \right) \right) \lor \neg b \overset{\ref{1.1.1}, \ref{1.1.5}}{=} \left( \neg a \lor b \right) \lor \neg b \end{equation*} \begin{equation*} \overset{\ref{2.10}}{=} \neg a \lor \left( b \lor \neg b \right) \overset{\ref{1.1.8}}{=} \neg a \lor 1 \overset{\ref{2.7}}{=} 1 \end{equation*} With \ref{1.1.7} and \ref{1.1.8} and \ref{2.3} we get: $$\neg \left(a \land b \right) = \neg a \lor \neg b$$ By the duality principle follows the second part. $\Box$
Notice that you proved that each element has a unique complement without using the associative law.
Given that each element has one and only one complement, in order to prove that $$\neg (a \wedge b) = \neg a \vee \neg b,$$ it suffices to show that $a \wedge b$ is a complement of $\neg a \vee \neg b$.
Using properties given in the definition, $$(a \wedge b) \wedge (\neg a \vee \neg b) = (a \wedge b \wedge \neg a) \vee (b \wedge \neg a \wedge \neg b) = 0 \vee 0 = 0,$$ and similarly $(a \wedge b) \vee (\neg a \vee \neg b) = 1$.
Hence $\neg (a \wedge b) = \neg a \vee \neg b$.
The equality $\neg (a \vee b) = \neg a \wedge \neg b$ follows too because the definitions are symmetric with respect to $\wedge$ and $\vee$.
Edit. Clearly, I used associativity in $a \wedge b \wedge \neg a = 0$. Let me try to get around it.
We certainly have that $(a \wedge b) \wedge \neg a = (b \wedge a) \wedge \neg a$. First, I'll add another auxiliary result:
Proof. $$y = (x \wedge \neg x) \vee y = (x \vee y) \wedge (\neg x \vee y) = x \wedge \neg x = 0.$$
Now, $$a \vee ( (b \wedge a) \wedge \neg a ) = (a \vee (b \wedge a)) \wedge (a \vee \neg a) = a \vee (b \wedge a) = a,$$ and similarly, $\neg a \vee ((b \wedge a) \wedge \neg a) = \neg a$, whence, by the above result, $(b \wedge a) \wedge \neg a = 0$.
Analogously, $b \wedge \neg a \wedge \neg b = 0$.