Theorem. $ f \: \text{1-1} \rightarrow f(A) \setminus f(B) = f(A\setminus B) $
Proof. $$ f \: \text{1-1} \rightarrow f(A) \setminus f(B) = f(A\setminus B) \leftrightarrow ( \underbrace{f(A) \setminus f(B) \subseteq f(A \setminus B)}_{(i)} \wedge \underbrace{f(A) \setminus f(B) \supseteq f(A \setminus B)}_{(ii)} ) $$ (i) Contrapositive: $$ f(x) = y \notin f(A \setminus B) \leftrightarrow y \notin f(A \setminus B) \wedge (x \in B \vee x \notin B) \leftrightarrow (\underbrace{y \notin f(A \setminus B) \wedge x \in B}_{s_1}) \vee (\underbrace{y \notin f(A \setminus B) \wedge x \notin B}_{s_2}) $$ For $y \notin f(A \setminus B)$ to hold, either $s_1$ or $s_2$ needs to be true.
$s_1$ implies $y \in f(B)$, since $y = f(x)$ and $x \in B$ which causes $y \in \{ f(x) : x \in B \}$.
$s_2$ implies $y \notin f(A)$, since $y \notin \{ f(x) : x \in A \wedge x \notin B \}$ and $x \notin B$.
Summarizing the above yields $$ y \notin f(A \setminus B) \leftrightarrow s_1 \vee s_2 \rightarrow y \in f(B) \vee y \notin f(A) \leftrightarrow \neg( y \in f(A) \wedge y \notin f(B) ) \leftrightarrow \neg(y \in f(A) \setminus f(B) ) \leftrightarrow y \notin f(A) \setminus f(B) $$ $ \Box $

(ii) Contrapositive: $$ y \notin (f(A) \setminus f(B)) \leftrightarrow y \notin f(A) \vee y \in f(B) $$ $f(x) = y \notin f(A) \supseteq f(A \setminus B)$ implies that $y \notin f(A \setminus B)$.
Given $f(x) = y \in f(B)$, injectivity of $f$ requires that $y \notin \{ f(x) : x \notin B \} \supseteq f(A \setminus B)$.
Hence, we have that $$ y \notin (f(A) \setminus f(B)) \rightarrow y \notin f(A \setminus B). $$ $\Box$

2014 DEC 24 (v2.0)
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