Htykuut

As the title says, is this true?

$$(lnot P to lnot Q) to (P to Q)$$

The truth table is

begin{array}{rrrrrr}
P & Q & lnot P & lnot Q & lnot P to lnot Q & P to Q & (lnot P to lnot Q) to (P to Q) \ hline
T & T & F & F & T & T & T \
T & F & F & T & T & F & F \
F & T & T & F & F & T & T \
F & F & T & T & T & T & T \
end{array}

It seems like it's true from the table.

If it is true, is it true because $$(lnot P to lnot Q) to (P to Q)$$ has the same truth table corresponding to the $to$ connective which is false only when the antecedent is T but the consequent is F?

Or is it true because the statement is true when the premises of $lnot P to lnot Q$ and $P to Q$ are true?

If it's not true, why not?

$(lnot Ptolnot Q)to(Pto Q)$ is not a tautology because it is not true when $P$ is true but $Q$ is false.   That is shown in the second row of your truth table.

Likewise, it is not a contradiction.   The statement is conditionally true.

The statement is logically equivalent to $lnot(Plandlnot Q)$, also to $(lnot Plor Q)$.

Now $(lnot Ptolnot Q)to(Qto P)$ is a tautology in classical logic.   Notice the order of the terms.

Indeed $lnot Pto lnot Q$ is the contrapositive of $Qto P$, and the two are logically equivalent.

No, if we have a statement "$P$ then $Q$", then "$neg P$ then $neg Q$" is the inverse of the statement. The inverse being true does not imply the statement is true.

For instance consider a class where the cutoff for an $A$ is $90%$. Consider the statement $$
text{"If you have above an }80%text{, then you will receive an }Atext{."}
$$
This statement is not true. However its inverse is true.
$$
text{"If you do not have above an }80%text{, then you will not receive an }Atext{."}
$$