For those who don't have the OP's cited source available, a Boolean algebra is initially defined to be a non-empty set $B$, on which two binary operations--called $\wedge$ and $\vee$--and a unary operation--called $'$--are all defined, where $B$ has two distinct elements $0,1$ (possibly among other elements), such that the following axioms are all satisfied:
$0'=1\text{ and }1'=0,\tag{11}$ $\forall p\in B[p\wedge 0=0\text{ and }p\vee 1=1],\tag{12}$ $\forall p\in B[p\wedge 1=p\text{ and }p\vee 0=p],\tag{13}$ $\forall p\in B[p\wedge p'=0\text{ and }p\vee p'=1],\tag{14}$ $\forall p\in B\left[(p')'=p\right],\tag{15}$ $\forall p\in B[p\wedge p=p\text{ and }p\vee p=p],\tag{16}$ $\forall p,q\in B\left[(p\wedge q)'=p'\vee q'\text{ and }(p\vee q)'=p'\wedge q'\right],\tag{17}$ $\forall p,q\in B[p\wedge q=q\wedge p\text{ and }p\vee q=q\vee p],\tag{18}$ $\forall p,q,r\in B\bigl[(p\wedge q)\wedge r=p\wedge(q\wedge r)\text{ and }(p\vee q)\vee r=p\vee(q\vee r)\bigr],\tag{19}$ $\forall p,q,r\in B\bigl[p\wedge (q\vee r)=(p\wedge q)\vee(p\wedge r)\text{ and }p\vee (q\wedge r)=(p\vee q)\wedge(p\vee r)\bigr].\tag{20}$
It is then claimed that $(13)$, $(14)$, $(18)$, and $(20)$ (and the assumptions about $B$) entail all the other axioms. Just for fun, I'm going to prove that we can make even fewer assumptions and still get all the desired properties.
Definition: We will say a Boolean algebra is a non-empty set $B$, on which two binary operations--called $\wedge$ and $\vee$--are defined, where $B$ has at least two distinct elements, such that the following axioms are all satisfied:
$\exists u,z\in B\forall p\in B\bigl[[p\wedge u=p\text{ and }p\vee z=p]\text{ and }\exists q\in B[p\wedge q=z\text{ and }p\vee q=u]\bigr],\tag{i}$ $\forall p,q\in B[p\wedge q=q\wedge p\text{ and }p\vee q=q\vee p],\tag{ii}$ $\forall p,q,r\in B\bigl[p\wedge (q\vee r)=(p\wedge q)\vee(p\wedge r)\text{ and }p\vee (q\wedge r)=(p\vee q)\wedge(p\vee r)\bigr].\tag{iii}$
Observe that the latter two axioms are simply $(18)$ and $(20)$, while the former is entailed by $(13)$ and $(14)$ together with the extra assumptions about $B$ (such as $0,1\in B$, and the existence of the unary operation $'$ on $B$).
Proposition A: The $u$ and $z$ from (i) are unique, and so is the $q$.
Proof: Suppose $u,v\in B$ are such that for all $p\in B$, $p\wedge u=p$ and $p\wedge v=p$. In particular, then, it follows from $(18)$ that $v=v\wedge u=u\wedge v=u,$ so uniqueness holds. We may similarly prove the uniqueness of $z$. Hereinafter, we refer to $z$ and $u$ as $0$ and $1$, respectively. Then $(13)$ holds, and by (i), we have $\forall p\in B\exists q\in B[p\wedge q=0\text{ and }p\vee q=1],$ a weakening of $(14)$. Once we've demonstrated that such a $q$ is uniquely determined by $p$, then we will be able to explicitly define the unary operation $'$ on $B$ so that $(14)$ is satisfied completely.
Take any $p\in B$, and suppose that $q,r\in B$ are such that $p\wedge q=0=p\wedge r$ and $p\vee q=1=p\vee r$. Then by $(13)$, $(18)$ and $(20)$, we see that $\begin{align}q &= q\wedge 1\\ &= q\wedge(p\vee r)\\ &= (q\wedge p)\vee(q\wedge r)\\ &= (r\wedge q)\vee(p\wedge q)\\ &= (r\wedge q)\vee 0\\ &= (r\wedge q)\vee(p\wedge r)\\ &= (r\wedge p)\vee(r\wedge q)\\ &= r\wedge(p\vee q)\\ &= r\wedge 1\\ &=r,\end{align}$ as desired.
Hereinafter, given $p\in B$, we will define $p'$ to be the unique $q\in B$ such that $p\wedge q=0$ and $p\vee q=1$. $\Box$
Lemma 1: $(11)$, $(12)$, and $(16)$ hold.
Proof: By $(14)$, we have $0'=0'\vee 0=1$. Other part of $(11)$ similar.
By $(13)$, $(18)$, and $(20)$, we have $p\vee 1=(p\vee 1)\wedge 1=(p\vee 1)\wedge(p\vee p')=p\vee(1\wedge p')=p\vee(p'\wedge 1)=p\vee p'=1.$ Other part of $(12)$ similar.
By $(13)$, $(14)$, and $(20)$, we have$p\wedge p=(p\wedge p)\vee 0=(p\wedge p)\vee(p\wedge p')=p\wedge(p\vee p')=p\wedge 1=p.$ Other part of $(16)$ similar. $\Box$
Lemma 2: $p\wedge(p\vee q)=p$ and $p\vee(p\wedge q)=p$.
Proof: By $(12)$, $(13)$, $(18)$, and $(20)$, we have $p\wedge(p\vee q)=(p\vee 0\wedge(p\vee q)=p\vee(0\wedge q)=p\vee(q\wedge 0)=p\vee 0=p.$ Other part similar. $\Box$
Lemma 3: $p\vee q=q$ if and only if $p\wedge q=p$.
Proof: If $p\vee q=q$, then by Lemma 2, $p=p\wedge(p\vee q)=p\wedge q$. Other direction similar. $\Box$
Proposition B: $(19)$ holds.
Proof: Let $x=p\vee(q\vee r)$ and $y=(p\vee q)\vee r$.
By Lemma 2--$p\wedge(p\vee q)=p$ portion--it follows that $p\wedge x=p$. By $(20)$ and by Lemma 2, we also have $\begin{align}p\wedge y &= p\wedge\bigl((p\vee q)\vee r\bigr)\\ &= \bigl(p\wedge(p\vee q)\bigr)\vee(p\wedge r)\\ &= p\vee(p\wedge r)\\ &= p,\end{align}$ so $p\vee x=p\vee y$.
Now, by $(13)$, $(14)$, $(18)$, and $(20)$, we have $\begin{align}p'\wedge x &= p'\wedge\bigl(p\vee (q\vee r)\bigr)\\ &= (p'\wedge p)\vee\bigl(p'\wedge(q\vee r)\bigr)\\ &= \bigl(p'\wedge(q\vee r)\bigr)\vee(p\wedge p')\\ &= \bigl(p'\wedge(q\vee r)\bigr)\vee 0\\ &= p'\wedge(q\vee r),\end{align}$ and also $\begin{align}p'\wedge y &= p'\wedge\bigl((p\vee q)\vee r\bigr)\\ &= \bigl(p'\wedge(p\vee q)\bigr)\vee(p'\wedge r)\\ &= \bigl((p'\wedge p)\vee(p'\wedge q)\bigr)\vee(p'\wedge r)\\ &= \bigl((p'\wedge q)\vee(p\wedge p')\bigr)\vee(p'\wedge r)\\ &= \bigl((p'\wedge q)\vee 0\bigr)\vee(p'\wedge r)\\ &= \bigl(p'\wedge q)\vee(p'\wedge r)\\ &= p'\wedge(q\vee r),\end{align}$ so $p'\wedge x=p'\wedge y$.
Finally, by $(13)$, $(14)$, $(18)$, $(20)$, and the identities derived above, we have $\begin{align}p\vee(q\vee r) &= x\\ &= x\wedge 1\\ &= x\wedge(p\vee p')\\ &= (x\wedge p)\vee(x\wedge p')\\ &= (p\wedge x)\vee(p'\wedge x)\\ &= (p\wedge y)\vee(p'\wedge y)\\ &= (y\wedge p)\vee(y\wedge p')\\ &= y\wedge(p\vee p')\\ &= y\wedge 1\\ &= y\\ &= (p\vee q)\vee r,\end{align}$ as desired. Other part of $(19)$ similar. $\Box$
Lemma 4: Define a relation $\leq$ on $B$ by $p\leq q$ iff $p\vee q=q$ iff $p\wedge q=p$. Then $\leq$ partially orders $B.$ Moreover, by $(12)$ $0,1$ are respectively the least and greatest elements of $B$ under the partial ordering by $\leq$, so since $B$ has at least $2$ elements by definition, then we have that $0\neq 1$.
Proof: Reflexivity holds by $(16)$, and antisymmetry holds by $(18)$. If $p\leq q$ and $q\leq r$, then by $(19)$ we have $p\vee r=p\vee(q\vee r)=(p\vee q)\vee r=q\vee r=r,$ so $p\leq r$, and so transitivity holds. Thus, $\leq$ partially orders $B$. $\Box$
Lemma 5: If $r\leq p,q$, then $r\leq p\wedge q$, and if $r\geq p,q$, then $r\geq p\vee q$. Also, by Lemma 2, $p\wedge q\leq p\leq p\vee q$ for all $p,q$. Thus, $p\wedge q$ and $p\vee q$ are (respectively) the greatest lower bound and least upper bound of $p,q$ in $B$ under the partial order $\leq$.
Proof: Suppose $r\leq p,q$, so $r\vee p=p$ and $r\vee q=q$, so $r\vee(p\wedge q)=(r\vee p)\wedge(r\vee q)=p\wedge q,$ and so $r\leq p\wedge q$. Other part similar. $\Box$
Lemma 6: If $p\leq q$, then $p\vee r\leq q\vee r$ and $p\wedge r\leq q\wedge r$ for all $r$.
Proof: If $p\leq q$, then $p\vee q=q$, so $(p\wedge r)\vee(q\wedge r)=(p\vee q)\wedge r=q\wedge r,$ and so $p\wedge r\leq q\wedge r$. Other part similar. $\Box$
Lemma 7: $p\vee q=1$ if and only if $p\geq q'$. $p\wedge q=0$ if and only if $p\leq q'$.
Proof: If $p\geq q'$, then by Lemma 6, we have $1\geq p\vee q\geq q'\vee q=1,$ so $p\vee q=1$ by antisymmetry. On the other hand, if $p\vee q=1$, then $p\vee q'=1\wedge(p\vee q')=(p\vee q)\wedge(p\vee q')=p\vee(q\wedge q')=p\vee 0=p,$ so $p\geq q'$. Other part similar. $\Box$
Proposition C: $(15)$ and $(17)$ hold.
Proof: Since $p\vee p'=1$, then by Lemma 7, we have $p\geq (p')'$. On the other hand, $p\wedge p'=0$, then $p\leq (p')'$. By antisymmetry, we have $p=(p')'$.
By $(12)$, $(14)$, $(16)$, $(18)$, $(19)$, and $(20)$, we have $\begin{align}(p'\wedge q')\vee(p\vee q) &= (p\vee q)\vee(p'\wedge q')\\ &= \bigl((p\vee q)\vee p'\bigr)\wedge\bigl((p\vee q)\vee q'\bigr)\\ &= \bigl((q\vee p)\vee p'\bigr)\wedge\bigl((p\vee q)\vee q'\bigr)\\ &= \bigl(q\vee (p\vee p')\bigr)\wedge\bigl(p\vee (q\vee q')\bigr)\\ &= \bigl(q\vee 1\bigr)\wedge\bigl(p\vee 1\bigr)\\ &= 1\wedge 1\\ &= 1,\end{align}$ so $p'\wedge q'\geq (p\vee q)'$ by Lemma 7, but on the other hand, $\begin{align}(p'\wedge q')\wedge(p\vee q) &= \bigl((p'\wedge q')\wedge p\bigr)\vee\bigl((p'\wedge q')\wedge q\bigr)\\ &= \bigl((q'\wedge p')\wedge p\bigr)\vee\bigl((p'\wedge q')\wedge q\bigr)\\ &= \bigl(q'\wedge (p'\wedge p)\bigr)\vee\bigl(p'\wedge (q'\wedge q)\bigr)\\ &= \bigl(q'\wedge (p\wedge p')\bigr)\vee\bigl(p'\wedge (q\wedge q')\bigr)\\ &= \bigl(q'\wedge 0\bigr)\vee\bigl(p'\wedge 0\bigr)\\ &= 0\wedge 0\\ &= 0,\end{align}$ so $p'\wedge q'\leq (p\vee q)'$ by Lemma 7, and so $p'\wedge q'=(p\vee q)'$ by antisymmetry.
Other part of $(17)$ similar. $\Box$