Double of Antiperiod is Period

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

Then $f$ is also periodic with a period of $2A$.

Proof
Let $L_f$ be the set of all periodic elements of $f$.

By Periodic Element is Multiple of Antiperiod and Absolute Value of Real Number is not less than Divisors:
 * $\forall p \in L_f: A \divides p \land A \le p$

Suppose there is a $p \in L_f$ such that $p = A$.

Then $\map f x = \map f {x + A} = - \map f x \implies \map f x = 0$, which contradicts Constant Function has no Period.

Therefore $\forall p \in L_f: A \divides p \land A \lt p$.

The smallest number $r$ such that $A \lt r \land A \divides r$ is $2A$.

But by Double of Antiperiodic Element is Periodic, this is a periodic element of $f$.

Hence the result.