If you are curious, the Wikipedia article Eight queens puzzle mentions
many aspects of the problem that go much beyond solving the
original puzzle.
The key idea in solving the puzzle using Boolean logic
variables is the identification of all of the constraints
needed to be satisfied for a valid placement solution.
The first constraint $\,Q_1\,$ is a conjunction of
disjunctions which require for every row that there is
at least one queen in that row.
All of the other constraints are of the form that
we forbid queen placements on certain pairs of $\,(i,j)\,$
locations. That is, we forbid a queen at $\,(i,j)\,$ and
a queen at $\,(k,l)\,$ for certain pairs of locations.
This is written as $\,\neg p(i,j) \vee \neg p(k,l)\,$
which is the negation of $\,p(i,j) \wedge p(k,l).\,$
The constraint $\,Q_2\,$ is to forbid
placing two queens in the same row. That is we forbid $\,p(i,j) \wedge p(i,k)\,$ for any $\,i,j,k\,$ although we only require $\,j<k.\,$ Similarly, $\,Q_3\,$ is to forbid
placing two queens in the same column. This condition is
$\,p(i,j) \wedge p(k,j)\,$ and we only require $\,i<k.\,$
The constraints $\,Q_4\,$ and $\,Q_5\,$ are to forbid
placing two queens in the same forward diagonal and the
same backward diagonal respectively.
In order to visualize how these constraints are written
in the $\,p(i,j) \wedge p(k,l) ,$ form, I suggest that
you try some explicit $\,i,j,k\,$ numbers and see how
they relate spatially on a chessboard grid.
I will show for 2 indices instead of 9, as it is simpler to see. But firstly I want to say that in Mathematics often symbols are very very ugly, but once you decompose what each part means it's quite simple.
$$\displaystyle\bigwedge\limits_{i=1}^2\bigwedge\limits_{n=1}^2\bigvee\limits_{j=1}^2p(i,j,n)$$
Let us decompose this! Maybe in this form, it will already be obvious what this means.
$$
\bigwedge\limits_{i=1}^2\left(\bigwedge\limits_{n=1}^2\left( \bigvee\limits_{j=1}^2p(i,j,n) \right)\right)
$$
If not, then let us decompose it, first we evaluate the inner parentheses:
$$
\bigwedge\limits_{i=1}^2\left(\bigwedge\limits_{n=1}^2[ p(i,1,n) \vee p(i,2,n) ]\right)
$$
Then we evaluate the second parentheses:
$$
\bigwedge\limits_{i=1}^2 \Big( [ p(i,1,1) \vee p(i,2,1) ] \wedge [ p(i,1,2) \vee p(i,2,2) ] \Big)
$$
Then finally we examine the last statement:
$$
\Big( [ p(1,1,1) \vee p(1,2,1) ] \wedge [ p(1,1,2) \vee p(1,2,2) ] \Big) \wedge \Big([ p(2,1,1) \vee p(2,2,1) ] \wedge [ p(2,1,2) \vee p(2,2,2) ] \Big)
$$
If these expansions confuse you, a trick I used when I started with math several years ago is just imagine you have all the terms specified by the subscripts and superscripts, then 'combine' them with the operation given (the large symbol on the left, for example, $\Sigma$ means after having all the terms, we add all of them). Of course there are (not so) minor details like commutativity/ability to re-arrange terms or not etc.
Best Answer
Yes, it’s a conjunction: $\bigwedge$ bears the same relationship to $\land$ as $\sum$ bears to $+$. Thus, for example, $\bigwedge_{i=1}^3p_i$ means exactly the same thing as $p_1\land p_2\land p_3$.
In the double conjunction $\bigwedge_{i=1}^{n-1}\bigwedge_{j=i+1}^n$ each value of $i$ from $1$ through $n-1$ is paired with each value of $j$ from $i+1$ through $n$; this has the effect of running through all pairs $\langle i,j\rangle$ of indices with $1\le i<j\le n$, so
$$\bigwedge_{i=1}^{n-1}\bigwedge_{j=i+1}^n(\neg p_i\lor\neg p_j)$$
is precisely equivalent to
$$\bigwedge_{1\le i<j\le n}(\neg p_i\lor\neg p_j)\,.\tag{1}$$
If $n=4$, for instance, this is
$$\begin{align*} &(\neg p_1\lor\neg p_2)\land(\neg p_1\lor\neg p_3)\land(\neg p_1\lor\neg p_4)\\ &\land(\neg p_2\lor\neg p_3)\land(\neg p_2\lor\neg p_4)\land(\neg p_3\lor\neg p_4)\,. \end{align*}$$
The net effect is to take the conjunction of all pairs $\neg p_i\lor\neg p_j$ for distinct $p_i$ and $p_j$.
The conjunction $(1)$ is true precisely when all of the disjunctions $\neg p_i\lor\neg p_j$ with $1\le i<j\le n$ are true; if even one of them is false, $(1)$ is also false.
Suppose that $1\le i<j\le n$, and $p_i$ and $p_j$ are both true. Then $\neg p_i$ and $\neg p_j$ are both false, so $\neg p_i\lor\neg p_j$ is false, and therefore $(1)$ is also false. In other words, if two or more of the propositions $p_1,\ldots,p_n$ are true, then $(1)$ is false. Now suppose that at most one of these propositions is true. If $1\le i<j\le n$, then at least one of $p_i$ and $p_j$ is false, so at least one of $\neg p_i$ and $\neg p_j$ is true, and therefore $\neg p_i\lor\neg p_j$ is true. Thus, each of the disjunctions $\neg p_i\lor\neg p_j$ is true, and therefore their conjunction $(1)$ is true. Thus, we have shown that $(1)$ is true if and only if at most one of the propositions $p_1,\ldots,p_n$ is true.