Enderton. Continue actually verifying set theorems.
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@ -1456,7 +1456,7 @@
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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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\code*{Bookshelf/Enderton/Set/Basic}
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\code*{Common/Set/Basic}
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{Set.not\_mem\_symm\_diff\_inter\_or\_not\_union}
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\begin{proof}
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@ -1948,7 +1948,7 @@
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\end{align*}
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\end{proof}
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\subsection{\pending{Exercise 2.15a}}%
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\subsection{\verified{Exercise 2.15a}}%
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\hyperlabel{sub:exercise-2.15a}
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Show that $A \cap (B + C) = (A \cap B) + (A \cap C)$.
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@ -1956,6 +1956,9 @@
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\lean*{Mathlib/Data/Set/Basic}
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{Set.inter\_symmDiff\_distrib\_left}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_2\_15a}
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\begin{proof}
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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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@ -1984,13 +1987,15 @@
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\end{align*}
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\end{proof}
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\subsection{\pending{Exercise 2.15b}}%
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\subsection{\verified{Exercise 2.15b}}%
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\hyperlabel{sub:exercise-2.15b}
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Show that $A + (B + C) = (A + B) + C$.
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\lean*{Mathlib/Data/Set/Basic}
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{Set.symmDiff\_assoc}
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\lean*{Mathlib/Order/SymmDiff}{symmDiff\_assoc}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_2\_15b}
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\begin{proof}
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Let $A$, $B$, and $C$ be sets.
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@ -2658,11 +2663,6 @@
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pairs $\pair{x, y}$ with $x \in A$ and $y \in B$.
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\end{theorem}
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\begin{note}
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The below Lean proof is a definition (i.e. an axiom).
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It does not prove such a set's existence from first principles.
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\end{note}
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\lean{Mathlib/SetTheory/ZFC/Basic}{Set.prod}
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\begin{proof}
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@ -2830,7 +2830,7 @@
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\hyperlabel{sub:lemma-1}
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\hyperlabel{sub:one-to-one-inverse}
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\begin{lemma}
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\begin{lemma}[1]
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For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\end{lemma}
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@ -3985,7 +3985,7 @@
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showing it can be represented as an ordered $1$-tuple as well.
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\end{answer}
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\subsection{\pending{Exercise 3.11}}%
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\subsection{\unverified{Exercise 3.11}}%
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\hyperlabel{sub:exercise-3.11}
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Prove the following version (for functions) of the extensionality principle:
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@ -6083,7 +6083,7 @@
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\section{Arithmetic}%
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\hyperlabel{sec:arithmetic}
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\subsection{\pending{Theorem 4I}}
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\subsection{\verified{Theorem 4I}}
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\hyperlabel{sub:theorem-4i}
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\begin{theorem}[4I]
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@ -6098,6 +6098,9 @@
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\lean{Init/Prelude}{Nat.add}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.theorem\_4i}
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\begin{proof}
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\paragraph{\eqref{sub:theorem-4i-eq1}}%
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@ -6115,7 +6118,7 @@
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\end{proof}
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\subsection{\pending{Theorem 4J}}
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\subsection{\verified{Theorem 4J}}
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\hyperlabel{sub:theorem-4j}
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\begin{theorem}[4J]
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@ -6130,6 +6133,9 @@
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\lean{Init/Prelude}{Nat.mul}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.theorem\_4j}
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\begin{proof}
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\paragraph{\eqref{sub:theorem-4j-eq1}}%
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@ -6145,7 +6151,7 @@
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\end{proof}
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\subsection{\pending{Left Additive Identity}}%
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\subsection{\verified{Left Additive Identity}}%
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\hyperlabel{sub:left-additive-identity}
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\begin{lemma}
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@ -6155,6 +6161,9 @@
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\lean{Init/Data/Nat/Basic}{Nat.zero\_add}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.left\_additive\_identity}
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\begin{proof}
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Let $S = \{n \in \omega \mid 0 + n = n\}$.
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@ -6185,17 +6194,20 @@
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\end{proof}
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\subsection{\pending{Lemma 3}}%
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\hyperlabel{sub:lemma-3}
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\subsection{\verified{Lemma 2}}%
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\hyperlabel{sub:lemma-2}
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\hyperlabel{sub:succ-add-eq-add-succ}
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\begin{lemma}
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\begin{lemma}[2]
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For all $m, n \in \omega$, $A_{m^+}(n) = A_m(n^+)$.
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In other words, $$m^+ + n = m + n^+.$$
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\end{lemma}
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\lean{Std/Data/Nat/Lemmas}{Nat.succ\_add\_eq\_succ\_add}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.lemma\_2}
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\begin{proof}
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Let $m \in \omega$ and define
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@ -6204,14 +6216,14 @@
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Afterwards we show that (iii) our theorem holds.
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\paragraph{(i)}%
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\hyperlabel{par:lemma-3-i}
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\hyperlabel{par:lemma-2-i}
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By \nameref{sub:theorem-4i}, $m^+ + 0 = m^+$.
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Likewise, $m + 0^+ = (m + 0)^+ = m^+$.
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Thus $0 \in S$.
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\paragraph{(ii)}%
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\hyperlabel{par:lemma-3-ii}
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\hyperlabel{par:lemma-2-ii}
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Suppose $n \in S$.
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By \nameref{sub:theorem-4i}, $m^+ + n^+ = (m^+ + n)^+$.
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@ -6221,13 +6233,13 @@
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\paragraph{(iii)}%
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By \nameref{par:lemma-3-i} and \nameref{par:lemma-3-ii}, $S$ is inductive.
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By \nameref{par:lemma-2-i} and \nameref{par:lemma-2-ii}, $S$ is inductive.
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Hence \nameref{sub:theorem-4b} implies $S = \omega$.
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Thus for all $n \in \omega$, $m^+ + n = m + n^+$.
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\end{proof}
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\subsection{\pending{Theorem 4K-1}}%
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\subsection{\verified{Theorem 4K-1}}%
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\hyperlabel{sub:theorem-4k-1}
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\begin{theorem}[4K-1]
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@ -6235,7 +6247,10 @@
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For $m, n, p \in \omega$, $$m + (n + p) = (m + n) + p.$$
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\end{theorem}
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\lean{Mathlib/Algebra/Group/Defs}{add\_assoc}
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\lean{Init/Data/Nat/Basic}{Nat.add\_assoc}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.theorem\_4k\_1}
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\begin{proof}
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@ -6281,7 +6296,7 @@
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\end{proof}
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\subsection{\pending{Theorem 4K-2}}%
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\subsection{\verified{Theorem 4K-2}}%
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\hyperlabel{sub:theorem-4k-2}
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\begin{theorem}[4K-2]
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@ -6289,7 +6304,10 @@
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For $m, n \in \omega$, $$m + n = n + m.$$
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\end{theorem}
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\lean{Mathlib/Algebra/Group/Defs}{add\_comm}
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\lean{Init/Data/Nat/Basic}{Nat.add\_comm}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.theorem\_4k\_2}
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\begin{proof}
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@ -6331,7 +6349,7 @@
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\end{proof}
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\subsection{\pending{Zero Multiplicand}}%
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\subsection{\verified{Zero Multiplicand}}%
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\hyperlabel{sub:zero-multiplicand}
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\begin{lemma}
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@ -6341,6 +6359,9 @@
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\lean{Init/Data/Nat/Basic}{Nat.zero\_mul}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.zero\_multiplicand}
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\begin{proof}
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Define
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\begin{align*}
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0 \cdot n^+
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& = 0 \cdot n + 0 & \textref{sub:theorem-4j} \\
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& = 0 + 0 & \eqref{sub:zero-multiplicand-eq1} \\
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& = 0. & \textref{ref:addition}
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& = 0 \cdot n & \textref{sub:theorem-4i} \\
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& = 0. & \eqref{sub:zero-multiplicand-eq1}
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\end{align*}
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Thus $n^+ \in S$.
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@ -6380,7 +6401,7 @@
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\end{proof}
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\subsection{\pending{Successor Distribution}}%
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\subsection{\verified{Successor Distribution}}%
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\hyperlabel{sub:successor-distribution}
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\begin{lemma}
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@ -6390,6 +6411,9 @@
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\lean{Init/Data/Nat/Basic}{Nat.succ\_mul}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.succ\_distrib}
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\begin{proof}
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Let $m \in \omega$ and define
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@ -6434,7 +6458,7 @@
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\end{proof}
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\subsection{\pending{Theorem 4K-3}}
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\subsection{\verified{Theorem 4K-3}}
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\hyperlabel{sub:theorem-4k-3}
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\begin{theorem}[4K-3]
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@ -6444,6 +6468,9 @@
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\lean{Init/Data/Nat/Basic}{Nat.left\_distrib}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.theorem\_4k\_3}
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\begin{proof}
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Fix $n, p \in \omega$ and define
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@ -6494,7 +6521,7 @@
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\end{proof}
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\subsection{\pending{Successor Identity}}%
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\subsection{\verified{Successor Identity}}%
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\hyperlabel{sub:successor-identity}
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\begin{lemma}
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\lean{Std/Data/Nat/Lemmas}{Nat.succ\_eq\_one\_add}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.succ\_identity}
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\begin{proof}
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Let
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@ -6545,7 +6575,7 @@
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\end{proof}
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\subsection{\pending{Right Multiplicative Identity}}%
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\subsection{\verified{Right Multiplicative Identity}}%
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\hyperlabel{sub:right-multiplicative-identity}
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\begin{lemma}
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@ -6553,7 +6583,10 @@
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In other words, $$m \cdot 1 = m.$$
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\end{lemma}
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\lean{Init/Data/Nat/Basic}{mul\_one}
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\lean{Init/Data/Nat/Basic}{Nat.mul\_one}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.right\_mul\_id}
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\begin{proof}
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\end{proof}
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\subsection{\pending{Theorem 4K-5}}
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\subsection{\verified{Theorem 4K-5}}
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\hyperlabel{sub:theorem-4k-5}
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\begin{theorem}[4K-5]
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two properties in the opposite direction.
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\end{note}
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\lean{Mathlib/Algebra/Group/Defs}{mul\_comm}
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\lean{Init/Data/Nat/Basic}{Nat.mul\_comm}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.theorem\_4k\_5}
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\begin{proof}
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\end{proof}
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\subsection{\pending{Theorem 4K-4}}%
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\subsection{\verified{Theorem 4K-4}}%
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\hyperlabel{sub:theorem-4k-4}
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\begin{theorem}[4K-4]
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For $m, n, p \in \omega$, $$m \cdot (n \cdot p) = (m \cdot n) \cdot p.$$
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\end{theorem}
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\lean{Mathlib/Algebra/Group/Defs}{mul\_assoc}
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\lean{Init/Data/Nat/Basic}{Nat.mul\_assoc}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.theorem\_4k\_4}
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\begin{proof}
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\section{Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}%
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\hyperlabel{sec:ordering-natural-numbers}
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\subsection{\pending{Ordering on Successor}}%
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\subsection{\unverified{Ordering on Successor}}%
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\hyperlabel{sub:ordering-successor}
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\begin{lemma}
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Thus $m \in n \iff m \in \omega \land m \in n$.
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\end{proof}
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\subsection{\pending{Lemma 4L(a)}}%
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\subsection{\unverified{Lemma 4L(a)}}%
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\hyperlabel{sub:lemma-4l-a}
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\begin{lemma}[4L(a)]
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Suppose $m^+ \in n^+$.
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The definition of \nameref{ref:successor} immediately implies that
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$m \in m^+$.
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By \nameref{sec:ordering-natural-numbers}, $m^+ \in n^+$ implies
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$m^+ \in n$ or $m^+ = n$.
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Likewise, $m^+ \in n^+$ implies $m^+ \in n$ or $m^+ = n$.
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If the latter, $m \in n$ immediately follows.
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If the former, we note $n$ is a transitive set by
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\nameref{sub:theorem-4f}.
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@ -6813,7 +6851,7 @@
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\end{proof}
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\subsection{\pending{Lemma 4L(b)}}%
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\subsection{\verified{Lemma 4L(b)}}%
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\hyperlabel{sub:lemma-4l-b}
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\begin{lemma}[4L(b)]
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@ -6822,6 +6860,9 @@
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\lean{Init/Prelude}{Nat.lt\_irrefl}
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\code{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.theorem\_4l\_b}
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\begin{proof}
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Define
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@ -1,6 +1,7 @@
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import Bookshelf.Enderton.Set.Chapter_1
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import Common.Set.Basic
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import Mathlib.Data.Set.Lattice
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import Mathlib.Order.SymmDiff
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/-! # Enderton.Set.Chapter_2
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@ -261,7 +262,7 @@ theorem anti_monotonicity_ii (A B : Set (Set α)) (h : A ⊆ B)
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#check Set.sInter_subset_sInter
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/-- #### ∩/- Associativity
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/-- #### Intersection/Difference Associativity
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Let `A`, `B`, and `C` be sets. Then `A ∩ (B - C) = (A ∩ B) - C`.
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-/
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@ -271,7 +272,8 @@ theorem inter_diff_assoc (A B C : Set α)
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_ = { x | x ∈ A ∧ (x ∈ B ∧ x ∉ C) } := rfl
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_ = { x | (x ∈ A ∧ x ∈ B) ∧ x ∉ C } := by
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ext _
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sorry
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simp only [Set.mem_setOf_eq]
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rw [and_assoc]
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_ = { x | x ∈ A ∩ B ∧ x ∉ C } := rfl
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_ = (A ∩ B) \ C := rfl
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@ -667,6 +669,87 @@ theorem exercise_2_14 : A \ (B \ C) ≠ (A \ B) \ C := by
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end
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/-- #### Exercise 2.15 (a)
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Show that `A ∩ (B + C) = (A ∩ B) + (A ∩ C)`.
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-/
|
||||
theorem exercise_2_15a (A B C : Set α)
|
||||
: A ∩ (B ∆ C) = (A ∩ B) ∆ (A ∩ C) := Eq.symm $
|
||||
calc (A ∩ B) ∆ (A ∩ C)
|
||||
_ = ((A ∩ B) \ (A ∩ C)) ∪ ((A ∩ C) \ (A ∩ B)) := rfl
|
||||
_ = ((A ∩ B) \ A) ∪
|
||||
((A ∩ B) \ C) ∪
|
||||
(((A ∩ C) \ A) ∪
|
||||
((A ∩ C) \ B)) := by
|
||||
iterate 2 rw [Set.diff_inter]
|
||||
_ = (A ∩ (B \ A)) ∪
|
||||
(A ∩ (B \ C)) ∪
|
||||
((A ∩ (C \ A)) ∪
|
||||
(A ∩ (C \ B))) := by
|
||||
iterate 4 rw [Set.inter_diff_assoc]
|
||||
_ = ∅ ∪ (A ∩ (B \ C)) ∪ (∅ ∪ (A ∩ (C \ B))) := by
|
||||
iterate 2 rw [Set.inter_diff_self]
|
||||
_ = (A ∩ (B \ C)) ∪ (A ∩ (C \ B)) := by
|
||||
simp only [Set.empty_union]
|
||||
_ = A ∩ ((B \ C) ∪ (C \ B)) := by
|
||||
rw [Set.inter_distrib_left]
|
||||
_ = A ∩ (B ∆ C) := rfl
|
||||
|
||||
#check Set.inter_symmDiff_distrib_left
|
||||
|
||||
/-- #### Exercise 2.15 (b)
|
||||
|
||||
Show that `A + (B + C) = (A + B) + C`.
|
||||
-/
|
||||
theorem exercise_2_15b (A B C : Set α)
|
||||
: A ∆ (B ∆ C) = (A ∆ B) ∆ C := by
|
||||
rw [Set.Subset.antisymm_iff]
|
||||
apply And.intro
|
||||
· show ∀ x, x ∈ A ∆ (B ∆ C) → x ∈ (A ∆ B) ∆ C
|
||||
intro x hx
|
||||
apply Or.elim hx
|
||||
· intro ⟨hA, nBC⟩
|
||||
rw [Set.not_mem_symm_diff_inter_or_not_union] at nBC
|
||||
apply Or.elim nBC
|
||||
· intro h
|
||||
have : x ∉ A ∆ B := Set.symm_diff_mem_both_not_mem hA h.left
|
||||
exact Set.symm_diff_mem_right this h.right
|
||||
· intro h
|
||||
have ⟨nB, nC⟩ : x ∉ B ∧ x ∉ C := not_or_de_morgan.mp h
|
||||
have : x ∈ A ∆ B := Set.symm_diff_mem_left hA nB
|
||||
exact Set.symm_diff_mem_left this nC
|
||||
· intro ⟨hx₁, hx₂⟩
|
||||
apply Or.elim hx₁
|
||||
· intro ⟨hB, nC⟩
|
||||
have : x ∈ A ∆ B := Set.symm_diff_mem_right hx₂ hB
|
||||
exact Set.symm_diff_mem_left this nC
|
||||
· intro ⟨hC, nB⟩
|
||||
have : x ∉ A ∆ B := Set.symm_diff_not_mem_both_not_mem hx₂ nB
|
||||
exact Set.symm_diff_mem_right this hC
|
||||
· show ∀ x, x ∈ (A ∆ B) ∆ C → x ∈ A ∆ (B ∆ C)
|
||||
intro x hx
|
||||
apply Or.elim hx
|
||||
· intro ⟨hAB, nC⟩
|
||||
apply Or.elim hAB
|
||||
· intro ⟨hA, nB⟩
|
||||
have : x ∉ B ∆ C := Set.symm_diff_not_mem_both_not_mem nB nC
|
||||
exact Set.symm_diff_mem_left hA this
|
||||
· intro ⟨hB, nA⟩
|
||||
have : x ∈ B ∆ C := Set.symm_diff_mem_left hB nC
|
||||
exact Set.symm_diff_mem_right nA this
|
||||
· intro ⟨hC, nAB⟩
|
||||
rw [Set.not_mem_symm_diff_inter_or_not_union] at nAB
|
||||
apply Or.elim nAB
|
||||
· intro ⟨hA, hB⟩
|
||||
have : x ∉ B ∆ C := Set.symm_diff_mem_both_not_mem hB hC
|
||||
exact Set.symm_diff_mem_left hA this
|
||||
· intro h
|
||||
have ⟨nA, nB⟩ : x ∉ A ∧ x ∉ B := not_or_de_morgan.mp h
|
||||
have : x ∈ B ∆ C := Set.symm_diff_mem_right nB hC
|
||||
exact Set.symm_diff_mem_right nA this
|
||||
|
||||
#check symmDiff_assoc
|
||||
|
||||
/-- #### Exercise 2.16
|
||||
|
||||
Simplify:
|
||||
|
|
|
|||
|
|
@ -19,7 +19,7 @@ namespace Enderton.Set.Chapter_3
|
|||
|
||||
If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`.
|
||||
-/
|
||||
theorem lemma_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
|
||||
lemma lemma_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
|
||||
: OrderedPair x y ∈ 𝒫 𝒫 C := by
|
||||
have hxs : {x} ⊆ C := Set.singleton_subset_iff.mpr hx
|
||||
have hxys : {x, y} ⊆ C := Set.mem_mem_imp_pair_subset hx hy
|
||||
|
|
|
|||
|
|
@ -18,6 +18,291 @@ theorem theorem_4c (n : ℕ)
|
|||
| 0 => simp
|
||||
| m + 1 => simp
|
||||
|
||||
#check Nat.exists_eq_succ_of_ne_zero
|
||||
|
||||
/-- #### Theorem 4I
|
||||
|
||||
For natural numbers `m` and `n`,
|
||||
```
|
||||
m + 0 = m,
|
||||
m + n⁺ = (m + n)⁺
|
||||
```
|
||||
-/
|
||||
theorem theorem_4i (m n : ℕ)
|
||||
: m + 0 = m ∧ m + n.succ = (m + n).succ := ⟨rfl, rfl⟩
|
||||
|
||||
/-- #### Theorem 4J
|
||||
|
||||
For natural numbers `m` and `n`,
|
||||
```
|
||||
m ⬝ 0 = 0,
|
||||
m ⬝ n⁺ = m ⬝ n + m .
|
||||
```
|
||||
-/
|
||||
theorem theorem_4j (m n : ℕ)
|
||||
: m * 0 = 0 ∧ m * n.succ = m * n + m := ⟨rfl, rfl⟩
|
||||
|
||||
/-- #### Left Additive Identity
|
||||
|
||||
For all `n ∈ ω`, `A₀(n) = n`. In other words, `0 + n = n`.
|
||||
-/
|
||||
lemma left_additive_identity (n : ℕ)
|
||||
: 0 + n = n := by
|
||||
induction n with
|
||||
| zero => simp
|
||||
| succ n ih =>
|
||||
calc 0 + n.succ
|
||||
_ = (0 + n).succ := rfl
|
||||
_ = n.succ := by rw [ih]
|
||||
|
||||
#check Nat.zero_add
|
||||
|
||||
/-- #### Lemma 2
|
||||
|
||||
For all `m, n ∈ ω`, `Aₘ₊(n) = Aₘ(n⁺)`. In other words, `m⁺ + n = m + n⁺`.
|
||||
-/
|
||||
lemma lemma_2 (m n : ℕ)
|
||||
: m.succ + n = m + n.succ := by
|
||||
induction n with
|
||||
| zero => rfl
|
||||
| succ n ih =>
|
||||
calc m.succ + n.succ
|
||||
_ = (m.succ + n).succ := rfl
|
||||
_ = (m + n.succ).succ := by rw [ih]
|
||||
_ = m + n.succ.succ := rfl
|
||||
|
||||
#check Nat.succ_add_eq_succ_add
|
||||
|
||||
/-- #### Theorem 4K-1
|
||||
|
||||
Associatve law for addition. For `m, n, p ∈ ω`,
|
||||
```
|
||||
m + (n + p) = (m + n) + p.
|
||||
```
|
||||
-/
|
||||
theorem theorem_4k_1 (m n p : ℕ)
|
||||
: m + (n + p) = (m + n) + p := by
|
||||
induction m with
|
||||
| zero => simp
|
||||
| succ m ih =>
|
||||
calc m.succ + (n + p)
|
||||
_ = m + (n + p).succ := by rw [lemma_2]
|
||||
_ = (m + (n + p)).succ := rfl
|
||||
_ = ((m + n) + p).succ := by rw [ih]
|
||||
_ = (m + n) + p.succ := rfl
|
||||
_ = (m + n).succ + p := by rw [lemma_2]
|
||||
_ = (m + n.succ) + p := rfl
|
||||
_ = (m.succ + n) + p := by rw [lemma_2]
|
||||
|
||||
#check Nat.add_assoc
|
||||
|
||||
/-- #### Theorem 4K-2
|
||||
|
||||
Commutative law for addition. For `m, n ∈ ω`,
|
||||
```
|
||||
m + n = n + m.
|
||||
```
|
||||
-/
|
||||
theorem theorem_4k_2 {m n : ℕ}
|
||||
: m + n = n + m := by
|
||||
induction m with
|
||||
| zero => simp
|
||||
| succ m ih =>
|
||||
calc m.succ + n
|
||||
_ = m + n.succ := by rw [lemma_2]
|
||||
_ = (m + n).succ := rfl
|
||||
_ = (n + m).succ := by rw [ih]
|
||||
_ = n + m.succ := by rfl
|
||||
|
||||
#check Nat.add_comm
|
||||
|
||||
/-- #### Zero Multiplicand
|
||||
|
||||
For all `n ∈ ω`, `M₀(n) = 0`. In other words, `0 ⬝ n = 0`.
|
||||
-/
|
||||
theorem zero_multiplicand (n : ℕ)
|
||||
: 0 * n = 0 := by
|
||||
induction n with
|
||||
| zero => simp
|
||||
| succ n ih =>
|
||||
calc 0 * n.succ
|
||||
_ = 0 * n + 0 := rfl
|
||||
_ = 0 * n := rfl
|
||||
_ = 0 := by rw [ih]
|
||||
|
||||
#check Nat.zero_mul
|
||||
|
||||
/-- #### Successor Distribution
|
||||
|
||||
For all `m, n ∈ ω`, `Mₘ₊(n) = Mₘ(n) + n`. In other words,
|
||||
```
|
||||
m⁺ ⬝ n = m ⬝ n + n.
|
||||
```
|
||||
-/
|
||||
theorem succ_distrib (m n : ℕ)
|
||||
: m.succ * n = m * n + n := by
|
||||
induction n with
|
||||
| zero => simp
|
||||
| succ n ih =>
|
||||
calc m.succ * n.succ
|
||||
_ = m.succ * n + m.succ := rfl
|
||||
_ = (m * n + n) + m.succ := by rw [ih]
|
||||
_ = m * n + (n + m.succ) := by rw [theorem_4k_1]
|
||||
_ = m * n + (n.succ + m) := by rw [lemma_2]
|
||||
_ = m * n + (m + n.succ) := by
|
||||
conv => left; arg 2; rw [theorem_4k_2]
|
||||
_ = (m * n + m) + n.succ := by rw [theorem_4k_1]
|
||||
_ = m * n.succ + n.succ := rfl
|
||||
|
||||
#check Nat.succ_mul
|
||||
|
||||
/-- #### Theorem 4K-3
|
||||
|
||||
Distributive law. For `m, n, p ∈ ω`,
|
||||
```
|
||||
m ⬝ (n + p) = m ⬝ n + m ⬝ p.
|
||||
```
|
||||
-/
|
||||
theorem theorem_4k_3 (m n p : ℕ)
|
||||
: m * (n + p) = m * n + m * p := by
|
||||
induction m with
|
||||
| zero => simp
|
||||
| succ m ih =>
|
||||
calc m.succ * (n + p)
|
||||
_ = m * (n + p) + (n + p) := by rw [succ_distrib]
|
||||
_ = m * (n + p) + n + p := by rw [← theorem_4k_1]
|
||||
_ = m * n + m * p + n + p := by rw [ih]
|
||||
_ = m * n + (m * p + n) + p := by rw [theorem_4k_1]
|
||||
_ = m * n + (n + m * p) + p := by
|
||||
conv => left; arg 1; arg 2; rw [theorem_4k_2]
|
||||
_ = (m * n + n) + (m * p + p) := by rw [theorem_4k_1, theorem_4k_1]
|
||||
_ = m.succ * n + m.succ * p := by rw [succ_distrib, succ_distrib]
|
||||
|
||||
/-- #### Successor Identity
|
||||
|
||||
For all `m ∈ ω`, `Aₘ(1) = m⁺`. In other words, `m + 1 = m⁺`.
|
||||
-/
|
||||
theorem succ_identity (m : ℕ)
|
||||
: m + 1 = m.succ := by
|
||||
induction m with
|
||||
| zero => simp
|
||||
| succ m ih =>
|
||||
calc m.succ + 1
|
||||
_ = m + (Nat.succ Nat.zero).succ := by rw [lemma_2]
|
||||
_ = (m + 1).succ := rfl
|
||||
_ = m.succ.succ := by rw [ih]
|
||||
|
||||
#check Nat.succ_eq_one_add
|
||||
|
||||
/-- #### Right Multiplicative Identity
|
||||
|
||||
For all `m ∈ ω`, `Mₘ(1) = m`. In other words, `m ⬝ 1 = m`.
|
||||
-/
|
||||
theorem right_mul_id (m : ℕ)
|
||||
: m * 1 = m := by
|
||||
induction m with
|
||||
| zero => simp
|
||||
| succ m ih =>
|
||||
calc m.succ * 1
|
||||
_ = m * 1 + 1 := by rw [succ_distrib]
|
||||
_ = m + 1 := by rw [ih]
|
||||
_ = m.succ := by rw [succ_identity]
|
||||
|
||||
#check Nat.mul_one
|
||||
|
||||
/-- #### Theorem 4K-5
|
||||
|
||||
Commutative law for multiplication. For `m, n ∈ ω`, `m ⬝ n = n ⬝ m`.
|
||||
-/
|
||||
theorem theorem_4k_5 (m n : ℕ)
|
||||
: m * n = n * m := by
|
||||
induction m with
|
||||
| zero => simp
|
||||
| succ m ih =>
|
||||
calc m.succ * n
|
||||
_ = m * n + n := by rw [succ_distrib]
|
||||
_ = n * m + n := by rw [ih]
|
||||
_ = n * m + n * 1 := by
|
||||
conv => left; arg 2; rw [← right_mul_id n]
|
||||
_ = n * (m + 1) := by rw [← theorem_4k_3]
|
||||
_ = n * m.succ := by rw [succ_identity]
|
||||
|
||||
#check Nat.mul_comm
|
||||
|
||||
/-- #### Theorem 4K-4
|
||||
|
||||
Associative law for multiplication. For `m, n, p ∈ ω`,
|
||||
```
|
||||
m ⬝ (n ⬝ p) = (m ⬝ n) ⬝ p.
|
||||
```
|
||||
-/
|
||||
theorem theorem_4k_4 (m n p : ℕ)
|
||||
: m * (n * p) = (m * n) * p := by
|
||||
induction p with
|
||||
| zero => simp
|
||||
| succ p ih =>
|
||||
calc m * (n * p.succ)
|
||||
_ = m * (n * p + n) := rfl
|
||||
_ = m * (n * p) + m * n := by rw [theorem_4k_3]
|
||||
_ = (m * n) * p + m * n := by rw [ih]
|
||||
_ = p * (m * n) + m * n := by rw [theorem_4k_5]
|
||||
_ = p.succ * (m * n) := by rw [succ_distrib]
|
||||
_ = (m * n) * p.succ := by rw [theorem_4k_5]
|
||||
|
||||
#check Nat.mul_assoc
|
||||
|
||||
/-- #### Lemma 4L(b)
|
||||
|
||||
No natural number is a member of itself.
|
||||
-/
|
||||
lemma lemma_4l_b (n : ℕ)
|
||||
: ¬ n < n := by
|
||||
induction n with
|
||||
| zero => simp
|
||||
| succ n ih =>
|
||||
by_contra nh
|
||||
rw [Nat.succ_lt_succ_iff] at nh
|
||||
exact absurd nh ih
|
||||
|
||||
#check Nat.lt_irrefl
|
||||
|
||||
/-- #### Lemma 10
|
||||
|
||||
For every natural number `n ≠ 0`, `0 ∈ n`.
|
||||
-/
|
||||
theorem zero_least_nat (n : ℕ)
|
||||
: 0 = n ∨ 0 < n := by
|
||||
by_cases h : n = 0
|
||||
· left
|
||||
rw [h]
|
||||
· right
|
||||
have ⟨m, hm⟩ := Nat.exists_eq_succ_of_ne_zero h
|
||||
rw [hm]
|
||||
exact Nat.succ_pos m
|
||||
|
||||
/-- #### Trichotomy Law for ω
|
||||
|
||||
For any natural numbers `m` and `n`, exactly one of the three conditions
|
||||
```
|
||||
m ∈ n, m = n, n ∈ m
|
||||
```
|
||||
holds.
|
||||
-/
|
||||
theorem trichotomy_law_for_nat
|
||||
: IsAsymm ℕ LT.lt ∧ IsTrichotomous ℕ LT.lt :=
|
||||
⟨instIsAsymmLtToLT, instIsTrichotomousLtToLTToPreorderToPartialOrder⟩
|
||||
|
||||
/-- #### Linear Ordering on ω
|
||||
|
||||
Relation
|
||||
```
|
||||
∈_ω = {⟨m, n⟩ ∈ ω × ω | m ∈ n}
|
||||
```
|
||||
is a linear ordering on `ω`.
|
||||
-/
|
||||
theorem linear_ordering_on_nat
|
||||
: IsStrictTotalOrder ℕ LT.lt := isStrictTotalOrder_of_linearOrder
|
||||
|
||||
/-- #### Exercise 4.1
|
||||
|
||||
Show that `1 ≠ 3` i.e., that `∅⁺ ≠ ∅⁺⁺⁺`.
|
||||
|
|
@ -105,41 +390,4 @@ theorem exercise_4_14 (n : ℕ)
|
|||
have : even n := ⟨q, hq'⟩
|
||||
exact absurd this h
|
||||
|
||||
/-- #### Lemma 10
|
||||
|
||||
For every natural number `n ≠ 0`, `0 ∈ n`.
|
||||
-/
|
||||
theorem zero_least_nat (n : ℕ)
|
||||
: 0 = n ∨ 0 < n := by
|
||||
by_cases h : n = 0
|
||||
· left
|
||||
rw [h]
|
||||
· right
|
||||
have ⟨m, hm⟩ := Nat.exists_eq_succ_of_ne_zero h
|
||||
rw [hm]
|
||||
exact Nat.succ_pos m
|
||||
|
||||
/-- #### Trichotomy Law for ω
|
||||
|
||||
For any natural numbers `m` and `n`, exactly one of the three conditions
|
||||
```
|
||||
m ∈ n, m = n, n ∈ m
|
||||
```
|
||||
holds.
|
||||
-/
|
||||
theorem trichotomy_law_for_nat
|
||||
: IsAsymm ℕ LT.lt ∧ IsTrichotomous ℕ LT.lt :=
|
||||
⟨instIsAsymmLtToLT, instIsTrichotomousLtToLTToPreorderToPartialOrder⟩
|
||||
|
||||
/-- #### Linear Ordering on ω
|
||||
|
||||
Relation
|
||||
```
|
||||
∈_ω = {⟨m, n⟩ ∈ ω × ω | m ∈ n}
|
||||
```
|
||||
is a linear ordering on `ω`.
|
||||
-/
|
||||
theorem linear_ordering_on_nat
|
||||
: IsStrictTotalOrder ℕ LT.lt := isStrictTotalOrder_of_linearOrder
|
||||
|
||||
end Enderton.Set.Chapter_4
|
||||
|
|
@ -156,6 +156,46 @@ theorem prod_nonempty_nonempty_imp_nonempty_prod {A : Set α} {B : Set β}
|
|||
|
||||
/-! ## Symmetric Difference -/
|
||||
|
||||
/--
|
||||
If `x ∈ A` and `x ∉ B`, then `x ∈ A ∆ B`.
|
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-/
|
||||
theorem symm_diff_mem_left {A B : Set α} (hA : x ∈ A) (hB : x ∉ B)
|
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: x ∈ A ∆ B := by
|
||||
left
|
||||
exact ⟨hA, hB⟩
|
||||
|
||||
/--
|
||||
If `x ∉ A` and `x ∈ B`, then `x ∈ A ∆ B`.
|
||||
-/
|
||||
theorem symm_diff_mem_right {A B : Set α} (hA : x ∉ A) (hB : x ∈ B)
|
||||
: x ∈ A ∆ B := by
|
||||
right
|
||||
exact ⟨hB, hA⟩
|
||||
|
||||
/--
|
||||
If `x ∈ A` and `x ∈ B`, then `x ∉ A ∆ B`.
|
||||
-/
|
||||
theorem symm_diff_mem_both_not_mem {A B : Set α} (hA : x ∈ A) (hB : x ∈ B)
|
||||
: x ∉ A ∆ B := by
|
||||
intro h
|
||||
apply Or.elim h
|
||||
· intro ⟨_, nB⟩
|
||||
exact absurd hB nB
|
||||
· intro ⟨_, nA⟩
|
||||
exact absurd hA nA
|
||||
|
||||
/--
|
||||
If `x ∉ A` and `x ∉ B`, then `x ∉ A ∆ B`.
|
||||
-/
|
||||
theorem symm_diff_not_mem_both_not_mem {A B : Set α} (nA : x ∉ A) (nB : x ∉ B)
|
||||
: x ∉ A ∆ B := by
|
||||
intro h
|
||||
apply Or.elim h
|
||||
· intro ⟨hA, _⟩
|
||||
exact absurd hA nA
|
||||
· intro ⟨hB, _⟩
|
||||
exact absurd hB nB
|
||||
|
||||
/--
|
||||
`x` is a member of the `symmDiff` of `A` and `B` **iff** `x ∈ A ∧ x ∉ B` or
|
||||
`x ∉ A ∧ x ∈ B`.
|
||||
|
|
|
|||
Loading…
Reference in New Issue