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. $\Box$