Barber Paradox/Resolution 2

Resolution
Let the only condition above be relaxed, and rewrite it as:
 * his task was to shave every man in the community who did not shave himself.

The initial premises would be coded:
 * $(1): \quad \forall x \in \mathbb U: \paren {\neg \map S x} \implies \map B x$
 * $(2): \quad \map B b \iff \map S b$

Thus it is not the case that:
 * $\map B x \implies \paren {\neg \map S x}$

and so the barber is allowed to shave at least one man who does shave himself, the barber himself necessarily being one such.