Question

Prove A-(BnC) = (A-B)U(A-C), explain with an example​

Answer 1

Answer:

Prove set equality by showing that for any element $x$, $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$.

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$.

$B = \lbrace0,\, 1 \rbrace$.

$C = \lbrace0,\, 2 \rbrace$.

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$.

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$.

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$:

Assume that $x \in (A \backslash (B \cap C))$. Thus, $x \in A$ and $x \not \in (B \cap C)$.

Since $x \not \in (B \cap C)$, either $x \not \in B$ or $x \not \in C$ (or both.)

• If $x \not \in B$, then combined with $x \in A$, $x \in (A \backslash B)$.
• Similarly, if $x \not \in C$, then combined with $x \in A$, $x \in (A \backslash C)$.

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$:

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$. Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

• If $x \in (A \backslash B)$, then $x \in A$ and $x \not \in B$. Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$, is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$.
• Otherwise, if $x \in A \backslash C$, then $x \in A$ and $x \not \in C$. Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$. Therefore, $x \in A \backslash (B \cap C)$.

Either way, $x \in A \backslash (B \cap C)$.

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$, as required.