Enderton (set). Continue refining ordering theorems/exercises.
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@ -6863,6 +6863,8 @@
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.zero\_least\_nat}
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\lean{Std/Data/Nat/Init/Lemmas}{Nat.pos\_of\_ne\_zero}
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\begin{proof}
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Let $$S = \{n \in \omega \mid n = 0 \lor 0 \in n\}.$$
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@ -7255,7 +7257,7 @@
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\end{proof}
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\subsection{\pending{%
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\subsection{\verified{%
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Well Ordering of \texorpdfstring{$\omega$}{Natural Numbers}}}%
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\hyperlabel{sub:well-ordering-natural-numbers}
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@ -7264,6 +7266,9 @@
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Then there is some $m \in A$ such that $m \ineq n$ for all $n \in A$.
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\end{theorem}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.well\_ordering\_nat}
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\lean{Mathlib/SetTheory/Ordinal/Basic}{WellOrder}
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\begin{note}
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@ -7295,16 +7300,13 @@
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\hyperlabel{par:well-ordering-natural-numbers-ii}
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Suppose $n \in S$.
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That is, for all $m \in n$, $m \in \omega - A$.
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Let $p$ be an arbitrary element of $A$.
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By the \nameref{sub:trichotomy-law-natural-numbers}, only one of the
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following holds:
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$$p \in n^+, \quad p = n^+, \quad n^+ \in p.$$
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It cannot be that $p \in n^+$ since $p \in \omega - A$ by
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\eqref{sub:well-ordering-natural-numbers-eq1}.
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But then $n^+ = p$ or $n^+ \in p$, implying that $n^+$ is a least element
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of $A$.
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Since $A$ does not have a least element, it must be that $n^+ \not\in A$.
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We want to prove that $$\forall m, m \in n^+ \Rightarrow m \not\in A.$$
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To this end, let $m \in \omega$ such that $m \in n^+$.
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By definition of the \nameref{ref:successor}, $m \in n$ or $m = n$.
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If the former, $n \in S$ implies $m \not\in A$.
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If the latter, it isn't possible for $n \in A$ since the
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\nameref{sub:trichotomy-law-natural-numbers} would otherwise imply
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$n$ is the least element of $A$, which is assumed to not exist.
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Hence $n^+ \in S$.
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\paragraph{Conclusion}%
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@ -7333,7 +7335,7 @@
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Therefore it isn't possible $f(n^+) \in f(n)$ for all $n \in \omega$.
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\end{proof}
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\subsection{\pending{%
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\subsection{\verified{%
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Strong Induction Principle for \texorpdfstring{$\omega$}{Natural Numbers}}}%
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\hyperlabel{sub:strong-induction-principle-natural-numbers}
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@ -7347,6 +7349,9 @@
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Then $A = \omega$.
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\end{theorem}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.strong\_induction\_principle\_nat}
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\begin{proof}
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For the sake of contradiction, suppose $\omega - A$ is a nonempty set.
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By \nameref{sub:well-ordering-natural-numbers}, there exists a least element
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@ -7861,12 +7866,15 @@
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Refer to \nameref{sub:theorem-4k-5}.
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\end{proof}
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\subsection{\pending{Exercise 4.17}}%
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\subsection{\verified{Exercise 4.17}}%
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\hyperlabel{sub:exercise-4.17}
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Prove that $m^{n+p} = m^n \cdot m^p$.
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\lean*{Data/Nat/Lemmas}{Nat.pow\_add}
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\code*{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.exercise\_4\_17}
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\lean{Data/Nat/Lemmas}{Nat.pow\_add}
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\begin{proof}
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@ -7929,15 +7937,97 @@
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Therefore $$\img{\in_\omega^{-1}}{\{7, 8\}} = \{6, 7\}.$$
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\end{proof}
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\subsection{\sorry{Exercise 4.19}}%
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\subsection{\verified{Exercise 4.19}}%
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\hyperlabel{sub:exercise-4.19}
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Prove that if $m$ is a natural number and $d$ is a nonzero number, then there
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exist numbers $q$ and $r$ such that $m = (d \cdot q) + r$ and $r$ is less than
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$d$.
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exist numbers $q$ and $r$ such that $m = (d \cdot q) + r$ and $r$ is less
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than $d$.
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\code*{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.exercise\_4\_19}
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\begin{proof}
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TODO
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Let $d \in \omega$ such that $d \neq 0$.
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Define
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\begin{equation}
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\hyperlabel{sub:exercise-4.18-eq1}
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S = \{m \in \omega \mid
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(\exists q, d \in \omega) m = (d \cdot q) + r \land r < d\}.
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\end{equation}
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We prove that $S$ is an \nameref{ref:inductive-set} by showing
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(i) $0 \in S$ and (ii) if $m \in S$, then $m^+ \in S$.
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Afterward we prove (iii) the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:exercise-4.19-i}
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Let $q = 0$ and $r = 0$.
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We note $r < d$ by \textref{sub:zero-least-natural-number}.
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Furthermore,
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\begin{align*}
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(d \cdot 0) + 0
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& = d \cdot 0 & \textref{sub:theorem-4i} \\
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& = 0. & \textref{sub:theorem-4j}
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\end{align*}
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Thus $0 \in S$.
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\paragraph{(ii)}%
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\hyperlabel{par:exercise-4.19-ii}
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Suppose $m \in S$.
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Then there exists $q, d \in \omega$ such that $m = (d \cdot q) + r$ and
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$r < d$.
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Then
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\begin{align*}
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m^+
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& = ((d \cdot q) + r)^+ \\
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& = (d \cdot q) + r^+. & \textref{sub:theorem-4i}
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\end{align*}
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By \nameref{sub:trichotomy-law-natural-numbers}, there are three cases to
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consider:
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\subparagraph{Case 1}%
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Suppose $r^+ \in d$.
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Then it immediately follows $m^+ \in S$.
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\subparagraph{Case 2}%
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Suppose $r^+ = d$.
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Then
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\begin{align*}
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m^+
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& = (d \cdot q) + r^+ \\
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& = (d \cdot q) + d \\
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& = d \cdot q^+ & \textref{sub:theorem-4j} \\
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& = (d \cdot q^+) + 0. & \textref{sub:theorem-4i}
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\end{align*}
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Hence $m^+ \in S$.
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\subparagraph{Case 3}%
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Suppose $d \in r^+$.
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Then, by definition of the \nameref{ref:successor}, $d \in r$ or
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$d = r$.
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But $r \in d$.
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Thus, by \nameref{sub:trichotomy-law-natural-numbers}, a contradiction
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is introduced.
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Hence this case is not possible.
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\subparagraph{Conclusion}%
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Since the above cases are exhaustive, it follows $m^+ \in S$.
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\paragraph{(iii)}%
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By \nameref{par:exercise-4.19-i} and \nameref{par:exercise-4.19-ii},
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$S \subseteq \omega$ is indeed an inductive set.
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By \nameref{sub:theorem-4b}, $S = \omega$.
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Thus for all $m, d \in \omega$ where $d \neq 0$, there exist numbers $q$
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and $r$ such that $m = (d \cdot q) + r$ and $r < d$.
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\end{proof}
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\subsection{\sorry{Exercise 4.20}}%
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@ -7950,44 +8040,148 @@
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 4.21}}%
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\subsection{\unverified{Exercise 4.21}}%
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\hyperlabel{sub:exercise-4.21}
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Show that no natural number is a subset of any of its elements.
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\begin{proof}
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TODO
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Let $n$ be a natural number.
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Suppose $m \in n$.
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By \nameref{sub:trichotomy-law-natural-numbers}, $n \neq m$ and
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$n \not\in m$.
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By \nameref{sub:corollary-4m}, $n \not\subseteq m$.
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\end{proof}
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\subsection{\sorry{Exercise 4.22}}%
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\subsection{\verified{Exercise 4.22}}%
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\hyperlabel{sub:exercise-4.22}
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Show that for any natural numbers $m$ and $p$ we have $m \in m + p^+$.
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\code*{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.exercise\_4\_22}
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\begin{proof}
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TODO
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Let $m$ be a natural number and $$S = \{p \in \omega \mid m \in m + p^+\}.$$
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We prove that (i) $0 \in S$ and (ii) if $p \in S$, then $p^+ \in S$.
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Afterward we prove (iii) the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:exercise-4.22-i}
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By definition of the \nameref{ref:successor}, $m^+ = m \cup \{m\}$.
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Thus $m \in m^+ = m + 1$ (by \nameref{sub:successor-identity}).
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Hence $0 \in S$.
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\paragraph{(ii)}%
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\hyperlabel{par:exercise-4.22-ii}
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Suppose $p \in S$.
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That is, $m \in m + p^+$.
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By definition of the \nameref{ref:successor},
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\begin{align*}
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m + p^+
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& \in (m + p^+) \cup \{m + p^+\} \\
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& = (m + p^+)^+ \\
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& = m + p^{++}. & \textref{sub:theorem-4i}.
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\end{align*}
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By \nameref{sub:theorem-4f}, $m + p^{++}$ is a
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\nameref{ref:transitive-set}.
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Therefore $m \in m + p^+ \in m + p^{++}$ implies $m \in m + p^{++}$.
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Hence $p^+ \in S$.
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\paragraph{(iii)}%
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By \nameref{par:exercise-4.22-i} and \nameref{par:exercise-4.22-ii},
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$S \subseteq \omega$ is an inductive set.
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Thus \nameref{sub:theorem-4b} implies $S = \omega$.
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Hence for any natural numbers $m$ and $p$, we have $m \in m + p^+$.
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\end{proof}
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\subsection{\sorry{Exercise 4.23}}%
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\subsection{\verified{Exercise 4.23}}%
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\hyperlabel{sub:exercise-4.23}
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Assume that $m$ and $n$ are natural numbers with $m$ less than $n$.
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Show that there is some $p$ in $\omega$ for which $m + p^+ = n$.
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(It follows form this and the preceding exercise that $m$ is less than $n$ iff
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(It follows from this and the preceding exercise that $m$ is less than $n$ iff
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$(\exists p \in \omega)m + p^+ = n$.)
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\code*{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.exercise\_4\_23}
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\begin{proof}
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TODO
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Let $$S = \{n \in \omega \mid
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(\forall m \in n, \exists p \in \omega) m + p^+ = n\}.$$
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We prove that (i) $0 \in S$ and (ii) if $p \in S$, then $p^+ \in S$.
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Afterward we prove (iii) the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:exercise-4.23-i}
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It vacuously holds that $0 \in S$.
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\paragraph{(ii)}%
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\hyperlabel{par:exercise-4.23-ii}
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Suppose $n \in S$.
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Let $m \in n^+$.
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By definition of the \nameref{ref:successor}, $m \in n$ or $m = n$.
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If $m \in n$, then there exists some $p \in \omega$ such that
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$m + p^+ = n$.
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But then \nameref{sub:theorem-4i} implies
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\begin{align*}
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n^+
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& = (m + p^+)^+ \\
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& = m + p^{++}.
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\end{align*}
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If instead $m = n$, then \nameref{sub:successor-identity} implies that
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$$n^+ = m^+ = m + 1 = m + 0^+.$$
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Hence $n^+ \in S$.
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\paragraph{(iii)}%
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By \nameref{par:exercise-4.23-i} and \nameref{par:exercise-4.23-ii}, $S$
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is an \nameref{ref:inductive-set}.
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By \nameref{sub:theorem-4b}, $S = \omega$.
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Thus for all $m, n \in \omega$ such that $m \in n$, there exists some
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$p \in \omega$ such that $m + p^+ = n$.
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\end{proof}
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\subsection{\sorry{Exercise 4.24}}%
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\subsection{\verified{Exercise 4.24}}%
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\hyperlabel{sub:exercise-4.24}
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Assume that $m + n = p + q$.
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Show that $$m \in p \iff n \in q.$$
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Show that $$m \in p \iff q \in n.$$
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\code*{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.exercise\_4\_24}
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\begin{proof}
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TODO
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Let $m, n, p, q \in \omega$ such that $m + n = p + q$.
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\paragraph{($\Rightarrow$)}%
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Suppose $m \in p$.
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By \nameref{sub:theorem-4n}, $m + n \in p + n$.
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By hypothesis, $m + n = p + q$ meaning $p + q \in p + n$.
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By \nameref{sub:theorem-4k-2}, $q + p \in n + p$.
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Therefore another application of \nameref{sub:theorem-4n} implies
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$q \in n$.
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\paragraph{($\Leftarrow$)}%
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Suppose $q \in n$.
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By \nameref{sub:theorem-4n}, $q + p \in n + p$.
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By \nameref{sub:theorem-4k-2}, $p + q \in p + n$.
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By hypothesis, $p + q = m + n$ meaning $m + n \in p + n$.
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Therefore another application of \nameref{sub:theorem-4n} implies
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$m \in p$.
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\end{proof}
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\subsection{\sorry{Exercise 4.25}}%
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@ -1,5 +1,7 @@
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import Common.Logic.Basic
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import Common.Set.Basic
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import Mathlib.Data.Set.Basic
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import Mathlib.SetTheory.Ordinal.Basic
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/-! # Enderton.Set.Chapter_4
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@ -280,6 +282,8 @@ theorem zero_least_nat (n : ℕ)
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rw [hm]
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exact Nat.succ_pos m
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#check Nat.pos_of_ne_zero
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/-! #### Theorem 4N
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For any natural numbers `n`, `m`, and `p`,
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@ -386,7 +390,52 @@ Let `A` be a nonempty subset of `ω`. Then there is some `m ∈ A` such that
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-/
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theorem well_ordering_nat (A : Set ℕ) (hA : Set.Nonempty A)
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: ∃ m ∈ A, ∀ n, n ∈ A → m ≤ n := by
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sorry
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-- Assume `A` does not have a least element.
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by_contra nh
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simp only [not_exists, not_and, not_forall, not_le, exists_prop] at nh
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-- If we show the complement of `A` is `ω`, then `A = ∅`, a contradiction.
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suffices A.compl = Set.univ by
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have h := Set.univ_diff_compl_eq_self A
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rw [this] at h
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simp only [sdiff_self, Set.bot_eq_empty] at h
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exact absurd h.symm (Set.Nonempty.ne_empty hA)
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-- Use strong induction to prove every element of `ω` is in the complement.
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have : ∀ n : ℕ, (∀ m, m < n → m ∈ A.compl) := by
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intro n
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induction n with
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| zero =>
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intro m hm
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exact False.elim (Nat.not_lt_zero m hm)
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| succ n ih =>
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intro m hm
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have hm' : m < n ∨ m = n := by
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rw [Nat.lt_succ] at hm
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exact Nat.lt_or_eq_of_le hm
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apply Or.elim hm'
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· intro h
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exact ih m h
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· intro h
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have : ∀ x : ℕ, x ∈ A → n ≤ x := by
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intro x hx
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exact match @trichotomous ℕ LT.lt _ n x with
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| Or.inl h₁ => Nat.le_of_lt h₁
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| Or.inr (Or.inl h₁) => Nat.le_of_eq h₁
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| Or.inr (Or.inr h₁) => False.elim (ih x h₁ hx)
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by_cases hn : n ∈ A
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· have ⟨p, hp⟩ := nh n hn
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exact absurd hp.left (ih p hp.right)
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· rw [h]
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exact hn
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ext x
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simp only [Set.mem_univ, iff_true]
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by_contra nh'
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have ⟨y, hy₁, hy₂⟩ := nh x (show x ∈ A from Set.not_not_mem.mp nh')
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exact absurd hy₁ (this x y hy₂)
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#check WellOrder
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/-- #### Strong Induction Principle for ω
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@ -394,9 +443,23 @@ Let `A` be a subset of `ω`, and assume that for every `n ∈ ω`, if every numb
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less than `n` is in `A`, then `n ∈ A`. Then `A = ω`.
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-/
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theorem strong_induction_principle_nat (A : Set ℕ)
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(h : ∀ n : ℕ, (∀ m : ℕ, m < n → m ∈ A) → n ∈ A)
|
||||
(h : ∀ n : ℕ, (∀ x : ℕ, x < n → x ∈ A) → n ∈ A)
|
||||
: A = Set.univ := by
|
||||
sorry
|
||||
suffices A.compl = ∅ by
|
||||
have h' := Set.univ_diff_compl_eq_self A
|
||||
rw [this] at h'
|
||||
simp only [Set.diff_empty] at h'
|
||||
exact h'.symm
|
||||
|
||||
by_contra nh
|
||||
have ⟨m, hm⟩ := well_ordering_nat A.compl (Set.nmem_singleton_empty.mp nh)
|
||||
refine absurd (h m ?_) hm.left
|
||||
|
||||
-- Show that every number less than `m` is in `A`.
|
||||
intro x hx
|
||||
by_contra nx
|
||||
have : x < x := Nat.lt_of_lt_of_le hx (hm.right x nx)
|
||||
simp at this
|
||||
|
||||
/-- #### Exercise 4.1
|
||||
|
||||
|
|
@ -485,4 +548,126 @@ theorem exercise_4_14 (n : ℕ)
|
|||
have : even n := ⟨q, hq'⟩
|
||||
exact absurd this h
|
||||
|
||||
/-- #### Exercise 4.17
|
||||
|
||||
Prove that `mⁿ⁺ᵖ = mⁿ ⬝ mᵖ.`
|
||||
-/
|
||||
theorem exercise_4_17 (m n p : ℕ)
|
||||
: m ^ (n + p) = m ^ n * m ^ p := by
|
||||
induction p with
|
||||
| zero => calc m ^ (n + 0)
|
||||
_ = m ^ n := rfl
|
||||
_ = m ^ n * 1 := by rw [right_mul_id]
|
||||
_ = m ^ n * m ^ 0 := rfl
|
||||
| succ p ih => calc m ^ (n + p.succ)
|
||||
_ = m ^ (n + p).succ := rfl
|
||||
_ = m ^ (n + p) * m := rfl
|
||||
_ = m ^ n * m ^ p * m := by rw [ih]
|
||||
_ = m ^ n * (m ^ p * m) := by rw [theorem_4k_4]
|
||||
_ = m ^ n * m ^ p.succ := rfl
|
||||
|
||||
/-- #### Exercise 4.19
|
||||
|
||||
Prove that if `m` is a natural number and `d` is a nonzero number, then there
|
||||
exist numbers `q` and `r` such that `m = (d ⬝ q) + r` and `r` is less than `d`.
|
||||
-/
|
||||
theorem exercise_4_19 (m d : ℕ) (hd : d ≠ 0)
|
||||
: ∃ q r : ℕ, m = (d * q) + r ∧ r < d := by
|
||||
induction m with
|
||||
| zero =>
|
||||
refine ⟨0, 0, ?_⟩
|
||||
simp only [Nat.zero_eq, mul_zero, add_zero, true_and]
|
||||
exact Nat.pos_of_ne_zero hd
|
||||
| succ m ih =>
|
||||
have ⟨q, r, hm, hr⟩ := ih
|
||||
have hm' := calc m.succ
|
||||
_ = ((d * q) + r).succ := by rw [hm]
|
||||
_ = (d * q) + r.succ := rfl
|
||||
match @trichotomous ℕ LT.lt _ r.succ d with
|
||||
| Or.inl h₁ =>
|
||||
exact ⟨q, r.succ, hm', h₁⟩
|
||||
| Or.inr (Or.inl h₁) =>
|
||||
refine ⟨q.succ, 0, ?_, Nat.pos_of_ne_zero hd⟩
|
||||
calc Nat.succ m
|
||||
_ = d * q + Nat.succ r := hm'
|
||||
_ = d * q + d := by rw [h₁]
|
||||
_ = d * q.succ := rfl
|
||||
_ = d * q.succ + 0 := rfl
|
||||
| Or.inr (Or.inr h₁) =>
|
||||
have : d < d := calc d
|
||||
_ ≤ r := Nat.lt_succ.mp h₁
|
||||
_ < d := hr
|
||||
simp at this
|
||||
|
||||
/-- #### Exercise 4.22
|
||||
|
||||
Show that for any natural numbers `m` and `p` we have `m ∈ m + p⁺`.
|
||||
-/
|
||||
theorem exercise_4_22 (m p : ℕ)
|
||||
: m < m + p.succ := by
|
||||
induction p with
|
||||
| zero => simp
|
||||
| succ p ih => calc m
|
||||
_ < m + p.succ := ih
|
||||
_ < m + p.succ.succ := Nat.lt.base (m + p.succ)
|
||||
|
||||
/-- #### Exercise 4.23
|
||||
|
||||
Assume that `m` and `n` are natural numbers with `m` less than `n`. Show that
|
||||
there is some `p` in `ω` for which `m + p⁺ = n`. (It follows from this and the
|
||||
preceding exercise that `m` is less than `n` iff (`∃p ∈ ω) m + p⁺ = n`.)
|
||||
-/
|
||||
theorem exercise_4_23 (m n : ℕ) (hm : m < n)
|
||||
: ∃ p : ℕ, m + p.succ = n := by
|
||||
induction n with
|
||||
| zero => simp at hm
|
||||
| succ n ih =>
|
||||
have : m < n ∨ m = n := by
|
||||
rw [Nat.lt_succ] at hm
|
||||
exact Nat.lt_or_eq_of_le hm
|
||||
apply Or.elim this
|
||||
· intro hm₁
|
||||
have ⟨p, hp⟩ := ih hm₁
|
||||
refine ⟨p.succ, ?_⟩
|
||||
exact Eq.symm $ calc n.succ
|
||||
_ = (m + p.succ).succ := by rw [← hp]
|
||||
_ = m + p.succ.succ := rfl
|
||||
· intro hm₁
|
||||
refine ⟨0, ?_⟩
|
||||
rw [hm₁]
|
||||
|
||||
/-- #### Exercise 4.24
|
||||
|
||||
Assume that `m + n = p + q`. Show that
|
||||
```
|
||||
m ∈ p ↔ q ∈ n.
|
||||
```
|
||||
-/
|
||||
theorem exercise_4_24 (m n p q : ℕ) (h : m + n = p + q)
|
||||
: m < p ↔ q < n := by
|
||||
apply Iff.intro
|
||||
· intro hm
|
||||
have hr : m + n < p + n := (theorem_4n_i m p n).mp hm
|
||||
rw [h] at hr
|
||||
conv at hr => left; rw [add_comm]
|
||||
conv at hr => right; rw [add_comm]
|
||||
exact (theorem_4n_i q n p).mpr hr
|
||||
· intro hq
|
||||
have hr : q + p < n + p := (theorem_4n_i q n p).mp hq
|
||||
conv at hr => left; rw [add_comm]
|
||||
conv at hr => right; rw [add_comm]
|
||||
rw [← h] at hr
|
||||
exact (theorem_4n_i m p n).mpr hr
|
||||
|
||||
/-- #### Exercise 4.25
|
||||
|
||||
Assume that `n ∈ m` and `q ∈ p`. Show that
|
||||
```
|
||||
(m ⬝ q) + (n ⬝ p) ∈ (m ⬝ p) + (n ⬝ q).
|
||||
```
|
||||
-/
|
||||
theorem exercise_4_25 (m n p q : ℕ) (h₁ : n < m) (h₂ : q < p)
|
||||
: (m * q) + (n * p) < (m * p) + (n * q) := by
|
||||
sorry
|
||||
|
||||
end Enderton.Set.Chapter_4
|
||||
|
|
@ -154,6 +154,24 @@ theorem prod_nonempty_nonempty_imp_nonempty_prod {A : Set α} {B : Set β}
|
|||
rw [← nonempty_iff_ne_empty, ← nonempty_iff_ne_empty]
|
||||
exact ⟨⟨a, ha⟩, ⟨b, hb⟩⟩
|
||||
|
||||
/-! ## Difference -/
|
||||
|
||||
/--
|
||||
For any set `A`, the difference between the sample space and `A` is the
|
||||
complement of `A`.
|
||||
-/
|
||||
theorem univ_diff_self_eq_compl (A : Set α) : Set.univ \ A = A.compl := by
|
||||
unfold Set.compl SDiff.sdiff instSDiffSet Set.diff
|
||||
simp only [mem_univ, true_and]
|
||||
|
||||
/--
|
||||
For any set `A`, the difference between the sample space and the complement of
|
||||
`A` is `A`.
|
||||
-/
|
||||
theorem univ_diff_compl_eq_self (A : Set α) : Set.univ \ A.compl = A := by
|
||||
unfold Set.compl SDiff.sdiff instSDiffSet Set.diff
|
||||
simp only [mem_univ, mem_setOf_eq, not_not, true_and, setOf_mem_eq]
|
||||
|
||||
/-! ## Symmetric Difference -/
|
||||
|
||||
/--
|
||||
|
|
|
|||
Loading…
Reference in New Issue