Euler Phi Function of Non-Square Semiprime/Proof 2

Proof
A semiprime with distinct prime factors is a square-free integer.


 * $\map \phi n = \ds \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$

where $p \divides n$ denotes the primes which divide $n$.

As there are $2$ prime factors: $p$ and $q$, this devolves to:
 * $\map \phi n = \paren {p - 1} \paren {q - 1}$

except when $p = 2$, in which case:
 * $\map \phi n = q - 1$

But when $p = 2$, we have that $p - 1 = 1$ and so:
 * $\paren {p - 1} \paren {q - 1} = q - 1$

Hence the result.