Enderton. Continue proofs/exercises of "Axioms and Operations."
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@ -112,6 +112,18 @@ For each formula $\phi$ not containing $B$, the following is an axiom:
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\end{axiom}
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\section{\defined{Symmetric Difference}}%
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\label{ref:symmetric-difference}
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The \textbf{symmetric difference} $A + B$ of sets $A$ and $B$ is the set
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$(A - B) \cup (B - A)$.
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\begin{definition}
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\lean*{Mathlib/Data/Set/Basic}{symmDiff\_def}
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\end{definition}
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\section{\defined{Union Axiom}}%
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\label{ref:union-axiom}
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@ -1404,10 +1416,64 @@ For any set $C$ and $\mathscr{A} \neq \emptyset$,
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\end{proof}
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\subsection{\verified{%
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\texorpdfstring{$\cap$/$-$}{Intersection/Difference} Associativity}}%
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\label{sub:intersection-difference-associativity}
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Let $A$, $B$, and $C$ be sets.
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Then $A \cap (B - C) = (A \cap B) - C$.
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\begin{proof}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_diff\_assoc}
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Let $A$, $B$, and $C$ be sets.
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By definition of the intersection and relative complement of sets,
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\begin{align*}
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A \cap (B - C)
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& = \{ x \mid x \in A \land x \in B - C \} \\
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& = \{ x \mid x \in A \land (x \in B \land x \not\in C) \} \\
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& = \{ x \mid (x \in A \land x \in B) \land x \not\in C \} \\
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& = \{ x \mid x \in A \cap B \land x \not \in C \} \\
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& = (A \cap B) - C.
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\end{align*}
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\end{proof}
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\subsection{\verified{Nonmembership of Symmetric Difference}}
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\label{sub:nonmembership-symmetric-difference}
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Let $A$ and $B$ be sets. $x \not\in A + B$ if and only if either
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$x \in A \cap B$ or $x \not\in A \cup B$.
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\begin{proof}
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\lean{Common/Set/Basic}{Set.not\_mem\_symm\_diff\_inter\_or\_not\_union}
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By definition of the \nameref{ref:symmetric-difference},
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\begin{align*}
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x \not\in A + B
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& = \neg(x \in A + B) \\
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& = \neg[x \in (A - B) \cup (B - A)] \\
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& = \neg[x \in (A - B) \lor x \in (B - A)] \\
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& = \neg[(x \in A \land x \not\in B) \lor
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(x \in B \land x \not\in A)] \\
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& = \neg(x \in A \land x \not\in B) \land
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\neg(x \in B \land x \not\in A) \\
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& = (x \not\in A \lor x \in B) \land (x \not\in B \lor x \in A) \\
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& = ((x \not\in A \lor x \in B) \land x \not\in B) \lor
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((x \not\in A \lor x \in B) \land x \in A) \\
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& = (x \not\in A \land x \not\in B) \lor (x \in B \land x \in A) \\
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& = \neg(x \in A \lor x \in B) \lor (x \in B \land x \in A) \\
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& = x \not\in A \cup B \text{ or } x \in A \cap B.
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\end{align*}
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\end{proof}
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\section{Exercises 4}%
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\label{sec:exercises-4}
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\subsection{\unverified{Exercise 4.11}}%
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\subsection{\verified{Exercise 4.11}}%
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\label{sub:exercise-4.11}
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Show that for any sets $A$ and $B$,
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@ -1416,11 +1482,60 @@ Show that for any sets $A$ and $B$,
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_11\_i}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_11\_ii}
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\noindent Let $A$ and $B$ be sets.
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We prove that
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\begin{enumerate}[(i)]
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\item $A = (A \cap B) \cup (A - B)$
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\item $A \cup (B - A) = A \cup B$
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\end{enumerate}
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\paragraph{(i)}%
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By definition of the intersection, union, and relative complements of sets,
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\begin{align*}
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(A \cap B) \cup (A - B)
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& = \{ x \mid x \in A \cap B \lor x \in A - B \} \\
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& = \{ x \mid x \in \{ y \mid y \in A \land y \in B \} \lor
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x \in A - B \} \\
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& = \{ x \mid (x \in A \land x \in B) \lor x \in A - B \} \\
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& = \{ x \mid (x \in A \land x \in B) \lor
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x \in \{ y \mid y \in A \land y \not\in B \} \} \\
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& = \{ x \mid (x \in A \land x \in B) \lor
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(x \in A \land x \not\in B) \} \\
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& = \{ x \mid x \in A \lor (x \in B \land x \not\in B) \} \\
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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 union and relative complements of sets,
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\begin{align*}
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A \cup (B - A)
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& = \{ x \mid x \in A \lor x \in B - A \} \\
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& = \{ x \mid x \in A \lor
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x \in \{ y \mid y \in B \land y \not\in A \} \} \\
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& = \{ x \mid x \in A \lor (x \in B \land x \not\in A) \} \\
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& = \{ x \mid (x \in A \lor x \in B) \land
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(x \in A \lor x \not\in A) \} \\
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& = \{ x \mid (x \in A \lor x \in B) \land T \} \\
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& = \{ x \mid x \in A \lor x \in B \} \\
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& = \{ x \mid x \in A \cup B \} \\
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& = A \cup B.
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\end{align*}
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\end{proof}
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\subsection{\unverified{Exercise 4.12}}%
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\subsection{\verified{Exercise 4.12}}%
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\label{sub:exercise-4.12}
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Verify the following identity (one of De Morgan's laws):
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@ -1428,22 +1543,22 @@ Verify the following identity (one of De Morgan's laws):
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\begin{proof}
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TODO
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Refer to \nameref{sub:de-morgans-laws}.
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\end{proof}
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\subsection{\unverified{Exercise 4.13}}%
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\subsection{\verified{Exercise 4.13}}%
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\label{sub:exercise-4.13}
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Show that if $A \subseteq B$, then $C - B \subseteq C - A$.
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\begin{proof}
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TODO
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Refer to \nameref{sub:anti-monotonicity}.
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\end{proof}
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\subsection{\unverified{Exercise 4.14}}%
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\subsection{\verified{Exercise 4.14}}%
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\label{sub:exercise-4.14}
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Show by example that for some sets $A$, $B$, and $C$, the set $A - (B - C)$ is
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@ -1451,35 +1566,157 @@ Show by example that for some sets $A$, $B$, and $C$, the set $A - (B - C)$ is
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_14}
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Let $A = \{1, 2, 3\}$, $B = \{2, 3, 4\}$, and $C = \{3, 4, 5\}$.
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Then
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\begin{align*}
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A - (B - C)
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& = \{1, 2, 3\} - (\{2, 3, 4\} - \{3, 4, 5\}) \\
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& = \{1, 2, 3\} - \{2\} \\
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& = \{1, 3\}
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\end{align*}
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but
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\begin{align*}
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(A - B) - C
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& = (\{1, 2, 3\} - \{2, 3, 4\}) - \{3, 4, 5\} \\
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& = \{1\} - \{3, 4, 5\} \\
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& = \{1\}.
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\end{align*}
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\end{proof}
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\subsection{\unverified{Exercise 4.15}}%
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\label{sub:exercise-4.15}
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Define the symmetric difference $A + B$ of sets $A$ and $B$ to be the set
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$(A - B) \cup (B - A)$.
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\subsubsection{\unverified{Exercise 4.15a}}%
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\label{ssub:exercise-4.15a}
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\subsection{\verified{Exercise 4.15a}}%
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\label{sub:exercise-4.15a}
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Show that $A \cap (B + C) = (A \cap B) + (A \cap C)$.
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\begin{proof}
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TODO
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_symmDiff\_distrib\_left}
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By definition of the intersection, \nameref{ref:symmetric-difference}, and
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relative complement of sets,
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\begin{align*}
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(A & \cap B) + (A \cap C) \\
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& = [(A \cap B) - (A \cap C)] \cup [(A \cap C) - (A \cap B)] \\
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& = [(A \cap B) - A] \\
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& \qquad \cup [(A \cap B) - C] \\
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& \qquad \cup [(A \cap C) - A] \\
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& \qquad \cup [(A \cap C) - B]
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& \textref{sub:de-morgans-laws} \\
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& = [A \cap (B - A)] \\
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& \qquad \cup [A \cap (B - C)] \\
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& \qquad \cup [A \cap (C - A)] \\
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& \qquad \cup [A \cap (C - B)]
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& \textref{sub:intersection-difference-associativity} \\
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& = \emptyset \\
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& \qquad \cup [A \cap (B - C)] \\
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& \qquad \cup \emptyset \\
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& \qquad \cup [A \cap (C - B)]
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& \textref{sub:identitives-involving-empty-set} \\
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& = [A \cap (B - C)] \cup [A \cap (C - B)] \\
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& = A \cap [(B - C) \cup (C - B)]
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& \textref{sub:distributive-laws} \\
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& = A \cap (B + C).
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\end{align*}
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\end{proof}
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\subsubsection{\unverified{Exercise 4.15b}}%
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\label{ssub:exercise-4.15b}
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\subsection{\verified{Exercise 4.15b}}%
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\label{sub:exercise-4.15b}
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Show that $A + (B + C) = (A + B) + C$.
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\begin{proof}
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TODO
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\lean{Mathlib/Data/Set/Basic}{Set.symmDiff\_assoc}
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\noindent Let $A$, $B$, and $C$ be sets.
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We prove that
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\begin{enumerate}[(i)]
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\item $A + (B + C) \subseteq (A + B) + C$
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\item $(A + B) + C \subseteq A + (B + C)$
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\end{enumerate}
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\paragraph{(i)}%
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\label{par:exercise-4.15b-i}
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Let $x \in A + (B + C)$.
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Then $x$ is in $A$ or in $B + C$, but not both.
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There are two cases to consider:
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\subparagraph{Case 1}%
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Suppose $x \in A$ and $x \not\in B + C$.
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Then, by \nameref{sub:nonmembership-symmetric-difference},
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(a) $x \in B \cap C$ or (b) $x \not\in B \cup C$.
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Suppose (a) was true.
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That is, $x \in B$ and $x \in C$.
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Since $x$ is a member of $A$ and $B$, $x \not\in (A + B)$.
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Since $x$ is not a member of $A + B$ but is a member of $C$,
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$x \in (A + B) + C$.
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Now suppose (b) was true.
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That is, $x \not\in B$ and $x \not\in C$.
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Since $x$ is a member of $A$ but not $B$, $x \in (A + B)$.
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Since $x$ is a member of $A + B$ but not $C$, $x \in (A + B) + C$.
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\subparagraph{Case 2}%
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Suppose $x \in B + C$ and $x \not\in A$.
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Then (a) $x \in B$ or (b) $x \in C$ but not both.
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Suppose (a) was true.
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That is, $x \in B$ and $x \not\in C$.
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Since $x$ is not a member of $A$ and is a member of $B$, $x \in A + B$.
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Since $x$ is a member of $A + B$ but not $C$, $x \in (A + B) + C$.
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Now suppose (b) was true.
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That is, $x \not\in B$ and $x \in C$.
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Since $x$ is not a member of $A$ nor a member of $B$, $x \not\in A + B$.
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Since $x$ is not a member of $A + B$ but is a member of $C$,
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$x \in (A + B) + C$.
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\paragraph{(ii)}%
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\label{par:exercise-4.15b-ii}
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Let $x \in (A + B) + C$.
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Then $x$ is in $A + B$ or in $C$, but not both.
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There are two cases to consider:
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\subparagraph{Case 1}%
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Suppose $x \in A + B$ and $x \not\in C$.
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Then (a) $x \in A$ or (b) $x \in B$ but not both.
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Suppose (a) was true.
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That is, $x \in A$ and $x \not\in B$.
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Since $x$ is not a member of $B$ nor $C$, $x \not\in B + C$.
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Since $x$ is not a member of $B + C$ but is a member of $A$,
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$x \in A + (B + C)$.
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Now Suppose (b) was true.
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That is, $x \not\in A$ and $x \in B$.
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Since $x$ is a member of $B$ and not of $C$, then $x \in B + C$.
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Since $x$ is a member of $B + C$ and not of $A$, $x \in A + (B + C)$.
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\subparagraph{Case 2}%
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Suppose $x \not\in A + B$ and $x \in C$.
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Then, by \nameref{sub:nonmembership-symmetric-difference},
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(a) $x \in A \cap B$ or (b) $x \not\in A \cup B$.
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Suppose (a) was true.
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That is, $x \in A \land x \in B$.
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Since $x$ is a member of $B$ and $C$, $x \not\in B + C$.
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Since $x$ is not a member of $B + C$ but is a member of $A$,
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$x \in A + (B + C)$.
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Now suppose (b) was true.
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That is, $x \not\in A$ and $x \not\in B$.
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Since $x$ is not a member of $B$ but is a member of $C$, $x \in B + C$.
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Since $x$ is a member of $B + C$ but not of $A$, $x \in A + (B + C)$.
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\paragraph{Conclusion}%
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In both \nameref{par:exercise-4.15b-i} and \nameref{par:exercise-4.15b-ii},
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the subcases are exhaustive and prove the desired subset relation.
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Therefore $A + (B + C) = (A + B) + C$.
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\end{proof}
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@ -1,7 +1,9 @@
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import Mathlib.Data.Set.Basic
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import Mathlib.Data.Set.Lattice
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import Mathlib.Tactic.LibrarySearch
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import Bookshelf.Enderton.Set.Chapter_1
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import Common.Logic.Basic
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import Common.Set.Basic
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/-! # Enderton.Chapter_2
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@ -198,7 +200,7 @@ theorem exercise_3_7b_ii
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apply Iff.intro
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· intro h
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by_contra nh
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rw [not_or] at nh
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rw [not_or_de_morgan] at nh
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have ⟨a, hA⟩ := Set.not_subset.mp nh.left
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have ⟨b, hB⟩ := Set.not_subset.mp nh.right
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rw [Set.ext_iff] at h
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@ -281,4 +283,139 @@ theorem exercise_3_10 (ha : a ∈ B)
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rw [← hb, Set.mem_setOf_eq]
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exact h₂
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/-- ### Exercise 4.11 (i)
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Show that for any sets `A` and `B`, `A = (A ∩ B) ∪ (A - B)`.
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-/
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theorem exercise_4_11_i {A B : Set α}
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: A = (A ∩ B) ∪ (A \ B) := by
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unfold Union.union Set.instUnionSet Set.union
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unfold SDiff.sdiff Set.instSDiffSet Set.diff
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unfold Inter.inter Set.instInterSet Set.inter
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unfold Membership.mem Set.instMembershipSet Set.Mem setOf
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simp only
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suffices ∀ x, (A x ∧ (B x ∨ ¬B x)) = A x by
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conv => rhs; ext x; rw [← and_or_left, this]
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intro x
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refine propext ?_
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apply Iff.intro
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· intro hx
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exact hx.left
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· intro hx
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exact ⟨hx, em (B x)⟩
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/-- ### Exercise 4.11 (ii)
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Show that for any sets `A` and `B`, `A ∪ (B - A) = A ∪ B`.
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-/
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theorem exercise_4_11_ii {A B : Set α}
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: A ∪ (B \ A) = A ∪ B := by
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unfold Union.union Set.instUnionSet Set.union
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unfold SDiff.sdiff Set.instSDiffSet Set.diff
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unfold Membership.mem Set.instMembershipSet Set.Mem setOf
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simp only
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suffices ∀ x, ((A x ∨ B x) ∧ (A x ∨ ¬A x)) = (A x ∨ B x) by
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conv => lhs; ext x; rw [or_and_left, this x]
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intro x
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refine propext ?_
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apply Iff.intro
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· intro hx
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exact hx.left
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· intro hx
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exact ⟨hx, em (A x)⟩
|
||||
|
||||
section
|
||||
|
||||
variable {A B C : Set ℕ}
|
||||
|
||||
variable {hA : A = {1, 2, 3}}
|
||||
variable {hB : B = {2, 3, 4}}
|
||||
variable {hC : C = {3, 4, 5}}
|
||||
|
||||
lemma right_diff_eq_insert_one_three : A \ (B \ C) = {1, 3} := by
|
||||
rw [hA, hB, hC]
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
unfold SDiff.sdiff Set.instSDiffSet Set.diff at hx
|
||||
unfold Membership.mem Set.instMembershipSet Set.Mem setOf at hx
|
||||
unfold insert Set.instInsertSet Set.insert setOf at hx
|
||||
have ⟨ha, hb⟩ := hx
|
||||
rw [not_and_de_morgan, not_or_de_morgan] at hb
|
||||
simp only [Set.mem_singleton_iff, not_not] at hb
|
||||
refine Or.elim ha Or.inl ?_
|
||||
intro hy
|
||||
apply Or.elim hb
|
||||
· intro hz
|
||||
exact Or.elim hy (absurd · hz.left) Or.inr
|
||||
· intro hz
|
||||
refine Or.elim hz Or.inr ?_
|
||||
intro hz₁
|
||||
apply Or.elim hy <;> apply Or.elim hz₁ <;>
|
||||
· intro hz₂ hz₃
|
||||
rw [hz₂] at hz₃
|
||||
simp at hz₃
|
||||
· intro hx
|
||||
unfold SDiff.sdiff Set.instSDiffSet Set.diff
|
||||
unfold Membership.mem Set.instMembershipSet Set.Mem setOf
|
||||
unfold insert Set.instInsertSet Set.insert setOf
|
||||
apply Or.elim hx
|
||||
· intro hy
|
||||
refine ⟨Or.inl hy, ?_⟩
|
||||
intro hz
|
||||
rw [hy] at hz
|
||||
unfold Membership.mem Set.instMembershipSet Set.Mem at hz
|
||||
unfold singleton Set.instSingletonSet Set.singleton setOf at hz
|
||||
simp only at hz
|
||||
· intro hy
|
||||
refine ⟨Or.inr (Or.inr hy), ?_⟩
|
||||
intro hz
|
||||
have hzr := hz.right
|
||||
rw [not_or_de_morgan] at hzr
|
||||
exact absurd hy hzr.left
|
||||
|
||||
lemma left_diff_eq_singleton_one : (A \ B) \ C = {1} := by
|
||||
rw [hA, hB, hC]
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
unfold SDiff.sdiff Set.instSDiffSet Set.diff at hx
|
||||
unfold Membership.mem Set.instMembershipSet Set.Mem setOf at hx
|
||||
unfold insert Set.instInsertSet Set.insert setOf at hx
|
||||
have ⟨⟨ha, hb⟩, hc⟩ := hx
|
||||
rw [not_or_de_morgan] at hb hc
|
||||
apply Or.elim ha
|
||||
· simp
|
||||
· intro hy
|
||||
apply Or.elim hy
|
||||
· intro hz
|
||||
exact absurd hz hb.left
|
||||
· intro hz
|
||||
exact absurd hz hc.left
|
||||
· intro hx
|
||||
refine ⟨⟨Or.inl hx, ?_⟩, ?_⟩ <;>
|
||||
· intro hy
|
||||
cases hy with
|
||||
| inl y => rw [hx] at y; simp at y
|
||||
| inr hz => cases hz with
|
||||
| inl y => rw [hx] at y; simp at y
|
||||
| inr y => rw [hx] at y; simp at y
|
||||
|
||||
/-- ### Exercise 4.14
|
||||
|
||||
Show by example that for some sets `A`, `B`, and `C`, the set `A - (B - C)` is
|
||||
different from `(A - B) - C`.
|
||||
-/
|
||||
theorem exercise_4_14 : A \ (B \ C) ≠ (A \ B) \ C := by
|
||||
rw [
|
||||
@right_diff_eq_insert_one_three A B C hA hB hC,
|
||||
@left_diff_eq_singleton_one A B C hA hB hC
|
||||
]
|
||||
intro h
|
||||
rw [Set.ext_iff] at h
|
||||
have := h 3
|
||||
simp at this
|
||||
|
||||
end
|
||||
|
||||
end Enderton.Set.Chapter_2
|
||||
|
|
@ -1,5 +1,6 @@
|
|||
import Common.Combinator
|
||||
import Common.Finset
|
||||
import Common.List
|
||||
import Common.Logic
|
||||
import Common.Real
|
||||
import Common.Set
|
||||
|
|
@ -0,0 +1 @@
|
|||
import Common.Logic.Basic
|
||||
|
|
@ -0,0 +1,18 @@
|
|||
import Mathlib.Logic.Basic
|
||||
import Mathlib.Tactic.Tauto
|
||||
|
||||
/-! # Common.Logic.Basic
|
||||
|
||||
Additional theorems and definitions related to basic logic.
|
||||
-/
|
||||
|
||||
/--
|
||||
The de Morgan law that distributes negation across a conjunction.
|
||||
-/
|
||||
theorem not_and_de_morgan {a b : Prop} : (¬(a ∧ b)) ↔ (¬ a ∨ ¬ b) := by
|
||||
tauto
|
||||
|
||||
/--
|
||||
Renaming of `not_or` to indicate its relationship to de Morgan's laws.
|
||||
-/
|
||||
theorem not_or_de_morgan : ¬(p ∨ q) ↔ ¬p ∧ ¬q := not_or
|
||||
|
|
@ -1,5 +1,7 @@
|
|||
import Mathlib.Data.Set.Basic
|
||||
|
||||
import Common.Logic.Basic
|
||||
|
||||
/-! # Common.Set.Basic
|
||||
|
||||
Additional theorems and definitions useful in the context of `Set`s.
|
||||
|
|
@ -87,4 +89,57 @@ theorem mem_mem_imp_pair_subset {x y : α}
|
|||
· intro hy'
|
||||
rwa [hy']
|
||||
|
||||
/-! ## Symmetric Difference -/
|
||||
|
||||
/--
|
||||
`x` is a member of the `symmDiff` of `A` and `B` **iff** `x ∈ A ∧ x ∉ B` or
|
||||
`x ∉ A ∧ x ∈ B`.
|
||||
-/
|
||||
theorem mem_symm_diff_iff_exclusive_mem {A B : Set α}
|
||||
: x ∈ (A ∆ B) ↔ (x ∈ A ∧ x ∉ B) ∨ (x ∉ A ∧ x ∈ B) := by
|
||||
unfold symmDiff
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
simp at hx
|
||||
conv => arg 2; rw [and_comm]
|
||||
exact hx
|
||||
· intro hx
|
||||
simp
|
||||
conv => arg 2; rw [and_comm]
|
||||
exact hx
|
||||
|
||||
/--
|
||||
`x` is not a member of the `symmDiff` of `A` and `B` **iff** `x ∈ A ∩ B` or
|
||||
`x ∉ A ∪ B`.
|
||||
|
||||
This is the contraposition of `mem_symm_diff_iff_exclusive_mem`.
|
||||
-/
|
||||
theorem not_mem_symm_diff_inter_or_not_union {A B : Set α}
|
||||
: x ∉ (A ∆ B) ↔ (x ∈ A ∩ B) ∨ (x ∉ A ∪ B) := by
|
||||
unfold symmDiff
|
||||
simp
|
||||
rw [
|
||||
not_or_de_morgan,
|
||||
not_and_de_morgan, not_and_de_morgan,
|
||||
not_not, not_not,
|
||||
not_or_de_morgan
|
||||
]
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
apply Or.elim hx.left
|
||||
· intro nA
|
||||
exact Or.elim hx.right
|
||||
(fun nB => Or.inr ⟨nA, nB⟩)
|
||||
(fun hA => absurd hA nA)
|
||||
· intro hB
|
||||
apply Or.elim hx.right
|
||||
(fun nB => absurd hB nB)
|
||||
(fun hA => Or.inl ⟨hA, hB⟩)
|
||||
· intro hx
|
||||
apply Or.elim hx
|
||||
· intro hy
|
||||
exact ⟨Or.inr hy.right, Or.inr hy.left⟩
|
||||
· intro hy
|
||||
exact ⟨Or.inl hy.left, Or.inl hy.right⟩
|
||||
|
||||
end Set
|
||||
Loading…
Reference in New Issue