Number of Friday 13ths in a Year

Theorem
In any given year, there are between $1$ and $3$ (inclusive) months in which the $13$th falls on a Friday.

Proof
The day of the week on which the $13$th falls is directy dependent upon the day of the week that the $1$st of the month falls.

From Months that Start on the Same Day of the Week, the months can be grouped into equivalence classes according to which day the month starts:

For a non-leap year, the set of equivalence classes is:
 * $\set {\set {\text {January}, \text {October} }, \set {\text {February}, \text {March}, \text {November} }, \set {\text {April}, \text {July} }, \set {\text {May} }, \set {\text {June} }, \set {\text {August} }, \set {\text {September}, \text {December} } }$

For a leap year, the set of equivalence classes is:
 * $\set {\set {\text {January}, \text {April}, \text {July} }, \set {\text {February}, \text {August} }, \set {\text {March}, \text {November} }, \set {\text {May} }, \set {\text {June} }, \set {\text {September}, \text {December} }, \set {\text {October} } }$

It can be seen that none of these equivalence classes has fewer than $1$ day, and none has more than $3$.

Hence the result.