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I'm trying to prove $A\setminus(B\setminus C) = (A\setminus B)\cup C$ using the fact that $C\subseteq A$. Somehow, I can use the relation $(A\setminus B)^c = A^c \cup B$. However, I can't really see how to start the proof.

Any help on how to start/structure this proof would be appreciated

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Such a proof is best done using both inclusions: prove that $A \setminus (B \setminus C) \subseteq (A \setminus B) \cup C$ and that $(A \setminus B) \cup C \subseteq A \setminus (B \setminus C)$.

  1. $A \setminus (B \setminus C) \subseteq (A \setminus B) \cup C$

    Let $x \in A \setminus (B \setminus C)$. Then $x \in A$ and $x \notin B\setminus C$. If $x \notin B \setminus C$, then $x \notin B$ (case 1) or $x \in C$ (case 2). So in case 1, $x \in A$ and $x \notin B$ hence $x \in A \setminus B$, and therefore $x \in (A \setminus B) \cup C$. In case $2$, $x \in C$ hence $x \in (A \setminus B) \cup C$. In both cases $x \in (A \setminus B) \cup C$, hence $A \setminus (B \setminus C) \subseteq (A \setminus B) \cup C$.

  2. Perhaps try this yourself in the same way as the first inclusion?

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  • $\begingroup$ For part 2. could you say: Let x ∈ (A \ B) U C. Then, x ∈ (A \ B) or x ∈ C. Therefore, x ∈ A, x ∈ B and x ∈ C, hence x ∈ A \ (B \ C)? $\endgroup$ – aL_eX Oct 26 '16 at 17:13
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    $\begingroup$ You can't deduce that $x$ is in all three of $A, B$ and $C$. You have to split up in two cases again: when $x$ is in $A\setminus B$ and when $x$ is in $C$. $\endgroup$ – TastyRomeo Oct 26 '16 at 17:53
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A funny way to prove set-theoretic equality is to use indicator functions. Given $A, B \subset X$, we have $\mathbb{1}_{A\setminus B} = \mathbb{1}_{A}(1 - \mathbb{1}_B)$. Then we have the following equalities:

$\mathbb{1}_{A\setminus (B\setminus C)} = \mathbb{1}_{A} - \mathbb{1}_{A}\mathbb{1}_{B} + \mathbb{1}_{A}\mathbb{1}_{B}\mathbb{1}_{C}$

$\mathbb{1}_{(A \setminus B) \cup C} = \mathbb{1}_{(A \setminus B)} + \mathbb{1}_{C} - \mathbb{1}_{(A \setminus B)\cap C} = \mathbb{1}_{A} - \mathbb{1}_{A}\mathbb{1}_{B} + \mathbb{1}_{C} - \mathbb{1}_{A}\mathbb{1}_{C} + \mathbb{1}_{A}\mathbb{1}_{B}\mathbb{1}_{C}$

Because $C \subset A$, $\mathbb{1}_{C} - \mathbb{1}_{A}\mathbb{1}_{C} = 0$, so that both indicator functions are equal.

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\begin{align} A-(B-C)&=A\cap (B\cap C^c)^c \\ &=A\cap (B^c\cup C) \\ &=(A\cap B^c)\cup (A\cap C) \\ &=(A-B)\cup C\tag1 \end{align} $(1)$ is for $$ C\subset A\implies A\cap C=C $$

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Using De Morgan's Laws in addition to the provided set rule:

$ A\setminus (B \setminus C) = A \setminus (( B \setminus C )^{c})^{c} = A \setminus (B^{c} \cup C)^{c} = A \setminus (B \cap C^{c}) =((A \setminus (B \cap C^{c}))^{c})^{c} = (A^{c} \cup (B \cap C^{c}))^{c} =( (A^{c} \cup B) \cap (A^{c} \cup C^{c}))^{c} = (A \setminus B) \cup C$ given $A \supseteq C $

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Does $C\subseteq A$ is an assumption? The identity is false. Take $A=B=\varnothing$ and $C=\{1,2,3\}$

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  • $\begingroup$ $C\subseteq A$ is explicitly stated. $\endgroup$ – Ennar Oct 26 '16 at 16:46
  • $\begingroup$ Yes, it is. Read the question. $\endgroup$ – Ennar Oct 26 '16 at 16:47
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    $\begingroup$ the title is misleading... $\endgroup$ – boaz Oct 26 '16 at 16:47
  • $\begingroup$ That is why the questions have bodies as well. $\endgroup$ – Ennar Oct 26 '16 at 16:48
  • $\begingroup$ The title was misleading. I edited it to include the crucial assumption. $\endgroup$ – Chris Culter Oct 26 '16 at 18:05

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