Why can $P_1 ⇔ P_2 ⇔ P_3 ⇔ \ldots ⇔ P_n$ be true when not all the $P$’s have the same truth value?
For example: If P1 = T P2 = T P3 = F P4 = F would this be true? T(P1) ⇔ T(P2) ⇔ F(P3) ⇔ F(P4) = T(P1&2) ⇔ F(P3) ⇔ F(P4) = F(P1,2,3) ⇔ F(P4) = True
If so how can an evaluation for a series of compound biconditionals with index n and x (individual truth values) be represented as a formula? Potentially statement is true with known x = 0; x = n; ?
How can P1 ⇔ P2 ⇔ P3, ... ⇔ Pn be True when P1 =/= P2 =/= P3 =/= Pn and all P are not false ?
Your ‘P1 =/= P2 =/= P3 =/= Pn’ is a little unclear, but from the title of your post, it is clear that your question is: how can $P1 ⇔ P2 ⇔ P3, ... ⇔ Pn$ be true if it is not the case that all $P$’s are true, nor the case that all $P$’s are false? In other words, how can $P1 ⇔ P2 ⇔ P3, ... ⇔ Pn$ be true if it is not the case that all $P$’s have the same truth-value? (I think this is what you tried to express with your ‘P1 =/= P2 =/= P3 =/= Pn’)
This is a good and important question, and many students express the same confusion when they run into this.
Here is the thing: the way the $\Leftrightarrow$ is used in mathematics as a meta-logical statement is different from how it is used as a logical truth-functional operator.
Specifically, as a metalogical statement we write $P_1 \Leftrightarrow P_2 \Leftrightarrow P_3 \Leftrightarrow P_4$ to mean that all these statements are equivalent to each other. That is, we have $P_1 \Leftrightarrow P_2$ and $P_1 \Leftrightarrow P_3$ and $P_1 \Leftrightarrow P_4$ and $P_2 \Leftrightarrow P_3$ and $P_2 \Leftrightarrow P_4$ and $P_3 \Leftrightarrow P_4$. And note, we have all those when we have $P_1 \Leftrightarrow P_2$ and $P_2 \Leftrightarrow P_3$ and $P_3 \Leftrightarrow P_4$. Indeed, we can see $P_1 \Leftrightarrow P_2 \Leftrightarrow P_3 \Leftrightarrow P_4$ as a notational short-cut for the latter. But note: used in this way, all the terms need to have the same truth-value.
However, if the $\Leftrightarrow$ is used as a logical truth-functional operator (what we call the bi-conditional), the $\Leftrightarrow$ works (somewhat surprisingly!) quite differently. In fact, as a binary truth-functional operator, we should first of all worry about whether it is even associative. That is, is it true that $P \Leftrightarrow (Q \Leftrightarrow R)$ is equivalent to $(P \Leftrightarrow Q) \Leftrightarrow R$? Well, if you do the truth-table, you will find that it is, and therefore we can drop parentheses and have general state,emnts like $P \Leftrightarrow Q \Leftrightarrow R$.
OK, but what would make such a generalized biconditional true? Well, as you just discovered, and contrary to your intuitions, $P_1 \Leftrightarrow P_2 \Leftrightarrow P_3 \Leftrightarrow P_4$ is not equivalent to $(P_1 \Leftrightarrow P_2) \land (P_2 \Leftrightarrow P_3) \land (P_3 \Leftrightarrow P_4)$. Instead, it turns out that a general biconditional is true if and only if an even number of its terms are false. Hence, you can both true and false terms, while the whole statement is still true.
This is a very practical reason to try and clearly separate between meta-logical equivalence and the truth-functional bi-conditional. Indeed, I personally really don’t like it when texts use the $\Leftrightarrow$ for the truth-functional operator. For that, I really like to see $\leftrightarrow$ being used, so that we can use $\Leftrightarrow$ just for meta-logical equivalence.
Finally, you ask: so if you cannot use $P_1 \leftrightarrow P_2 \leftrightarrow P_3 \leftrightarrow ... \leftrightarrow P_n$ to express that they all have the same truth-value, how then can you express that? Well, we can of course use the meta-logical expression $P_1 \Leftrightarrow P_2 \Leftrightarrow P_3 \Leftrightarrow ... \Leftrightarrow P_n$. But if you want a formula from formal logic, you'll have to do something like $(P_1 \leftrightarrow P_2) \land (P_2 \leftrightarrow P_3) \land ... \land (P_{n-1} \leftrightarrow P_n)$
