Question

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

Answer 1

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}`.

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 \\ot \in (B \cap C)`.

Since
`x \\ot \in (B \cap C)`, either
`x \\ot \in B` or
`x \\ot \in C` (or both.)

• If
  `x \\ot \in B`, then combined with
  `x \in A`,
  `x \in (A \backslash B)`.
• Similarly, if
  `x \\ot \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 \\ot \in B`. Notice that
  `(x \\ot \in B) \implies (x \\ot \in (B \cap C))` since the contrapositive of that statement,
  `(x \in (B \cap C)) \implies (x \in B)`, is true. Therefore,
  `x \\ot \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 \\ot \in C`. Similarly,
  `x \\ot \in C \!` implies
  `x \\ot \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.