Enderton. Continue proving laws in "Axioms and Operations."
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@ -897,49 +897,80 @@ For any sets $A$ and $B$,
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\paragraph{(i)}%
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By the definition of the union of sets,
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$A \cup B = \{ x \mid x \in A \lor x \in B \}$.
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Since $\lor$ is commutative, it follows
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\begin{align*}
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A \cup B
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& = \{ x \mid x \in A \lor x \in B \} \\
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& = \{ x \mid x \in B \lor x \in A \} \\
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& = B \cup A,
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& = B \cup A.
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\end{align*}
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where the last equality follows again from the definition of the union of
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sets.
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\paragraph{(ii)}%
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By the definition of the intersection of sets,
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$A \cap B = \{ x \mid x \in A \land x \in B \}$.
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Since $\land$ is commutative, it follows
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\begin{align*}
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A \cap B
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& = \{ x \mid x \in A \land x \in B \} \\
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& = \{ x \mid x \in B \land x \in A \} \\
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& = B \land A,
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& = B \land A.
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\end{align*}
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where the last equality follows again from the definition of the
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intersection of sets.
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\end{proof}
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\subsection{\unverified{Associative Laws}}%
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\subsection{\verified{Associative Laws}}%
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\label{sub:associative-laws}
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For any sets $A$, $B$ and $C$,
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\begin{align*}
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A \cup (B \cup C) & = (A \cup B) \cup C \\
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(A \cup B) \cup C & = A \cup (B \cup C)
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A \cap (B \cap C) & = (A \cap B) \cap C
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\end{align*}
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Mathlib/Data/Set/Basic}{Set.union\_assoc}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_assoc}
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Let $A$, $B$, and $C$ be sets.
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We show (i) $A \cup (B \cup C) = (A \cup B) \cup C$ and then (ii)
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$A \cap (B \cap C) = (A \cap B) \cap C$.
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\paragraph{(i)}%
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By the definition of the union of sets,
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\begin{align*}
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A \cup (B \cup C)
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& = \{ x \mid x \in A \lor x \in (B \cup C) \} \\
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& = \{ x \mid x \in A \lor x \in \{ y \mid y \in B \lor C \}\} \\
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& = \{ x \mid x \in A \lor (x \in B \lor x \in C) \} \\
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& = \{ x \mid (x \in A \lor x \in B) \lor x \in C \} \\
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& = \{ x \mid x \in \{ y \mid y \in A \lor y \in B \} \lor
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x \in C \} \\
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& = \{ x \mid x \in (A \cup B) \lor x \in C \} \\
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& = (A \cup B) \cup C.
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\end{align*}
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\paragraph{(ii)}%
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By the definition of the intersection of sets,
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\begin{align*}
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A \cap (B \cap C)
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& = \{ x \mid x \in A \land x \in (B \cap C) \} \\
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& = \{ x \mid x \in A \land
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x \in \{ y \mid y \in B \land y \in C \}\} \\
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& = \{ x \mid x \in A \land (x \in B \land x \in C) \} \\
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& = \{ x \mid (x \in A \land x \in B) \land x \in C \} \\
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& = \{ x \mid x \in \{ y \mid y \in A \land y \in B \} \land
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x \in C \} \\
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& = \{ x \mid x \in (A \cap B) \land x \in C \} \\
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& = (A \cap B) \cap C.
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\end{align*}
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\end{proof}
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\subsection{\unverified{Distributive Laws}}%
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\subsection{\verified{Distributive Laws}}%
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\label{sub:distributive-laws}
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For any sets $A$, $B$, and $C$,
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@ -950,11 +981,49 @@ For any sets $A$, $B$, and $C$,
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Mathlib/Data/Set/Basic}{Set.inter\_distrib\_left}
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\lean{Mathlib/Data/Set/Basic}{Set.union\_distrib\_left}
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Let $A$, $B$, and $C$ be sets.
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We show (i) $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ and then (ii)
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$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
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\paragraph{(i)}%
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By the definition of the union and intersection of sets,
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\begin{align*}
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A \cap (B \cup C)
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& = \{ x \mid x \in A \land x \in B \cup C \} \\
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& = \{ x \mid x \in A \land
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x \in \{ y \mid y \in B \lor y \in C \}\} \\
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& = \{ x \mid x \in A \land (x \in B \lor x \in C) \} \\
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& = \{ x \mid (x \in A \land x \in B) \lor
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(x \in A \land x \in C) \} \\
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& = \{ x \mid x \in A \cap B \lor x \in A \cap C \} \\
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& = (A \cap B) \cup (A \cap C).
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\end{align*}
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\paragraph{(ii)}%
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By the definition of the union and intersection of sets,
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\begin{align*}
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A \cup (B \cap C)
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& = \{ x \mid x \in A \lor x \in B \cap C \} \\
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& = \{ x \mid x \in A \lor
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x \in \{ y \mid y \in B \land y \in C \}\} \\
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& = \{ x \mid x \in A \lor (x \in B \land x \in C) \} \\
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& = \{ x \mid (x \in A \lor x \in B) \land
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(x \in A \lor x \in C) \} \\
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& = \{ x \mid x \in A \cup B \land x \in A \cup C \} \\
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& = (A \cup B) \cap (A \cup C).
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\end{align*}
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\end{proof}
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\subsection{\unverified{De Morgan's Laws}}%
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\subsection{\verified{De Morgan's Laws}}%
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\label{sub:de-morgans-laws}
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For any sets $A$, $B$, and $C$,
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@ -965,11 +1034,51 @@ For any sets $A$, $B$, and $C$,
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Mathlib/Data/Set/Basic}{Set.diff\_inter\_diff}
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\lean{Mathlib/Data/Set/Basic}{Set.diff\_inter}
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Let $A$, $B$, and $C$ be sets.
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We show (i) $C - (A \cup B) = (C - A) \cap (C - B)$ and then (ii)
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$C - (A \cap B) = (C - A) \cup (C - B)$.
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\paragraph{(i)}%
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By definition of the union, intersection, and relative complements of sets,
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\begin{align*}
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C - (A \cup B)
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& = \{ x \mid x \in C \land x \not\in A \cup B \} \\
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& = \{ x \mid x \in C \land
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x \not\in \{ y \mid y \in A \lor y \in B \}\} \\
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& = \{ x \mid x \in C \land \neg(x \in A \lor x \in B) \} \\
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& = \{ x \mid x \in C \land (x \not\in A \land x \not\in B) \} \\
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& = \{ x \mid (x \in C \land x \not\in A) \land
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(x \in C \land x \not\in B) \} \\
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& = \{ x \mid x \in (C - A) \land x \in (C - B) \} \\
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& = (C - A) \cap (C - B).
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\end{align*}
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\paragraph{(ii)}%
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By definition of the union, intersection, and relative complements of sets,
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\begin{align*}
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C - (A \cap B)
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& = \{ x \mid x \in C \land x \not\in A \cap B \} \\
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& = \{ x \mid x \in C \land
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x \not\in \{ y \mid y \in A \land y \in B \}\} \\
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& = \{ x \mid x \in C \land \neg(x \in A \land x \in B) \} \\
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& = \{ x \mid x \in C \land (x \not\in A \lor x \not\in B) \} \\
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& = \{ x \mid (x \in C \land x \not\in A) \lor
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(x \in C \land x \not\in B) \} \\
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& = \{ x \mid x \in (C - A) \lor x \in (C - B) \} \\
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& = (C - A) \cup (C - B).
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\end{align*}
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\end{proof}
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\subsection{\unverified{
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\subsection{\verified{%
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Identities Involving \texorpdfstring{$\emptyset$}{the Empty Set}}}%
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\label{sub:identitives-involving-empty-set}
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@ -982,7 +1091,55 @@ For any set $A$,
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Mathlib/Data/Set/Basic}{Set.union\_empty}
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\lean*{Mathlib/Data/Set/Basic}{Set.inter\_empty}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_diff\_self}
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Let $A$ be an arbitrary set. We prove (i) that $A \cup \emptyset = A$, (ii)
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$A \cap \emptyset = \emptyset$, and (iii) $A \cap (C - A) = \emptyset$.
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\paragraph{(i)}%
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By definition of the emptyset and union of sets,
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\begin{align*}
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A \cup \emptyset
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& = \{ x \mid x \in A \lor x \in \emptyset \} \\
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& = \{ x \mid x \in A \lor F \} \\
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& = \{ x \mid x \in A \} \\
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& = A.
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\end{align*}
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\paragraph{(ii)}%
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By definition of the emptyset and intersection of sets,
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\begin{align*}
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A \cap \emptyset
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& = \{ x \mid x \in A \land x \in \emptyset \} \\
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& = \{ x \mid x \in A \land F \} \\
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& = \{ x \mid F \} \\
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& = \{ x \mid x \neq x \} \\
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& = \emptyset.
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\end{align*}
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\paragraph{(iii)}%
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By definition of the emptyset, and the intersection and relative complement
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of sets,
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\begin{align*}
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A \cap (C - A)
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& = \{ x \mid x \in A \land x \in C - A \} \\
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& = \{ x \mid x \in A \land
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x \in \{ y \mid y \in C \land y \not\in A \}\} \\
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& = \{ x \mid x \in A \land (x \in C \land x \not\in A) \} \\
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& = \{ x \mid x \in C \land F \} \\
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& = \{ x \mid F \} \\
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& = \{ x \mid x \neq x \} \\
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& = \emptyset.
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\end{align*}
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\end{proof}
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