Explain why the following statement is true if and only if at most one value in the statement is true?

Solution 1:

If exactly one $p_k$ is true, then in each $\neg p_i\lor \neg p_j$ we can have at either $i=k$ or $j=k$, but not both. This is because it is guaranteed that always $i\ne j$ (in fact $i+1\le j\le n$). Hence at least one of $i,j$ is $\ne k$, hence at last one of $p_i,p_j$ is false, hence at least one of $\neg p_i,\neg p_j$ is true, hence $\neg p_i\lor \neg p_j$ is true. As this is the case for all considered $i,j$, the $\bigwedge$ of these is also true.

If two (or more) of the $p_k$ are true, let $i$ be minimal with $p_i$ true and $j$ maximal with $p_j$ true. Then $i<j$ (i.e., $i+1\le j\le n$) and the false $\neg p_i\lor \neg p_j$ makes $\bigwedge_{j=i+1}^n(\neg p_i\lor \neg p_j)$ and ultimately the whole statement false.

In order to find a formula that expresses that exactly one of the $p_k$ is true, observe that this is equivalent to $$ (\text{at most one $p_k$ is true})\land\neg(\text{all $p_k$ are false}).$$