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Case 1: $|A| = n\in\mathbb{Z}^+$

Since $B$ is uncountable, $|B| = 2^{\aleph_0}$.

Then $|B - A| = 2^{\aleph_0} - n = 2^{\aleph_0}$.

Therefore, $B - A$ is equinumerous to $B$, and hence $B - A$ is similar to $B$.

Case 2: $|A| = \aleph_0$

Again, we have $|B - A| = 2^{\aleph_0} - \aleph_0 = 2^{\aleph_0}$

Therefore, $B - A$ is equinumerous to $B$, and hence $B - A$ is similar to $B$.

Does this work?