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

Sagot :

jacob193

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.

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