Periodic Element is Multiple of Antiperiod

Theorem
Let $f: \R \to \R$ be a real anti-periodic function with anti-period $A$.

Let $L$ be a periodic element of $f$.

Then $A \divides L$.

Proof
Consider $A + L$:

And so $A + L$ is an anti-periodic element of $f$.

Combining Antiperiodic Element is Multiple of Antiperiod, Divides is Reflexive, and Common Divisor Divides Difference yields:
 * $A \divides \left({A + L}\right) \land A \divides A \implies A \divides L$

Hence the result.