Perfect Number is Primitive Semiperfect

Theorem
Let $n \in \Z_{>0}$ be a perfect number.

Then $n$ is also a primitive semiperfect number.

Proof
Let $n$ be perfect.

From Divisor of Perfect Number is Deficient, all divisors of $n$ are deficient.

But from Semiperfect Number is not Deficient, it follows that the divisors of $n$ cannot be semiperfect.

Hence the result, by definition of primitive semiperfect number.