Given any monoid (that is a set $A$ equipped with an associative operation $\cdot$ and an identity $1$), we can define "finite products" roughly by:
$$\prod_{i=1}^n a_i = a_1\cdot a_2\cdot \dots \cdot a_n$$
where $\prod_{i=1}^0 = 1$.
Possible monoids are for example $(\mathbb{R},\cdot, 1)$ yielding "$\prod$", $(\mathbb{R},+,0)$ yielding "$\sum$" or $(P(S), \cup, \emptyset)$ yielding "$\bigcup$" and so on and so forth.
So, we can also view a monoid as a set $A$ together with a map $A^* \to A, (a_n) \mapsto \prod_{i=1}^n a_i$ taking lists (words, tuples) of elements of $A$ to elements of $A$.
Occassionally however, we may find "maps" (broadly speaking) which not only accept finite lists, but also infinite lists or even bigger families of elements as objects.
For example, a complete lattice is a set $A$ equipped with maps $\bigvee$ and $\bigwedge$ taking abitrary families of elements of $A$ to elements of $A$.
Intuitively, if you take the set of all "small sets" (this is usually realized as a proper class) as the set $A$, then you get a complete lattice with operations $\bigcup$ and $\bigcap$ called union and intersection, which take families of elements of $A$ (that is sets of sets) to sets.
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.
Best Answer
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.