Answer:
Prove set equality by showing that for any element [tex]x[/tex], [tex]x \in (A \backslash (B \cap C))[/tex] if and only if [tex]x \in ((A \backslash B) \cup (A \backslash C))[/tex].
Example:
[tex]A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace[/tex].
[tex]B = \lbrace0,\, 1 \rbrace[/tex].
[tex]C = \lbrace0,\, 2 \rbrace[/tex].
[tex]\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}[/tex].
[tex]\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}[/tex].
Step-by-step explanation:
Proof for [tex][x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))][/tex] for any element [tex]x[/tex]:
Assume that [tex]x \in (A \backslash (B \cap C))[/tex]. Thus, [tex]x \in A[/tex] and [tex]x \not \in (B \cap C)[/tex].
Since [tex]x \not \in (B \cap C)[/tex], either [tex]x \not \in B[/tex] or [tex]x \not \in C[/tex] (or both.)
- If [tex]x \not \in B[/tex], then combined with [tex]x \in A[/tex], [tex]x \in (A \backslash B)[/tex].
- Similarly, if [tex]x \not \in C[/tex], then combined with [tex]x \in A[/tex], [tex]x \in (A \backslash C)[/tex].
Thus, either [tex]x \in (A \backslash B)[/tex] or [tex]x \in (A \backslash C)[/tex] (or both.)
Therefore, [tex]x \in ((A \backslash B) \cup (A \backslash C))[/tex] as required.
Proof for [tex][x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))][/tex]:
Assume that [tex]x \in ((A \backslash B) \cup (A \backslash C))[/tex]. Thus, either [tex]x \in (A \backslash B)[/tex] or [tex]x \in (A \backslash C)[/tex] (or both.)
- If [tex]x \in (A \backslash B)[/tex], then [tex]x \in A[/tex] and [tex]x \not \in B[/tex]. Notice that [tex](x \not \in B) \implies (x \not \in (B \cap C))[/tex] since the contrapositive of that statement, [tex](x \in (B \cap C)) \implies (x \in B)[/tex], is true. Therefore, [tex]x \not \in (B \cap C)[/tex] and thus [tex]x \in A \backslash (B \cap C)[/tex].
- Otherwise, if [tex]x \in A \backslash C[/tex], then [tex]x \in A[/tex] and [tex]x \not \in C[/tex]. Similarly, [tex]x \not \in C \![/tex] implies [tex]x \not \in (B \cap C)[/tex]. Therefore, [tex]x \in A \backslash (B \cap C)[/tex].
Either way, [tex]x \in A \backslash (B \cap C)[/tex].
Therefore, [tex]x \in ((A \backslash B) \cup (A \backslash C))[/tex] implies [tex]x \in A \backslash (B \cap C)[/tex], as required.
