Organize custom proofs over Lean proofs.
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@ -1244,7 +1244,7 @@ The union of a finite collection of line segments in a plane.
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\section{Exercises 1.11}%
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\hyperlabel{sec:exercises-1-11}
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\subsection{\pending{Exercise 1.11.4}}%
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\subsection{\verified{Exercise 1.11.4}}%
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\hyperlabel{sub:exercise-1.11.4}
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Prove that the greatest-integer function has the properties indicated:
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@ -1254,8 +1254,7 @@ The union of a finite collection of line segments in a plane.
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$\floor{x + n} = \floor{x} + n$ for every integer $n$.
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% FIXUP: padding
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\code{Bookshelf/Apostol/Chapter\_1\_11}
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\code*{Bookshelf/Apostol/Chapter\_1\_11}
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{Apostol.Chapter\_1\_11.exercise\_4a}
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\begin{proof}
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@ -1277,8 +1276,7 @@ The union of a finite collection of line segments in a plane.
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-\floor{x} - 1 & \text{otherwise}.
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\end{cases}$
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% FIXUP: padding
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\code{Bookshelf/Apostol/Chapter\_1\_11}
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\code*{Bookshelf/Apostol/Chapter\_1\_11}
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{Apostol.Chapter\_1\_11.exercise\_4b\_1}
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\code{Bookshelf/Apostol/Chapter\_1\_11}
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@ -1321,8 +1319,7 @@ The union of a finite collection of line segments in a plane.
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$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
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% FIXUP: padding
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\code{Bookshelf/Apostol/Chapter\_1\_11}
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\code*{Bookshelf/Apostol/Chapter\_1\_11}
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{Apostol.Chapter\_1\_11.exercise\_4c}
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\begin{proof}
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@ -1362,26 +1359,24 @@ The union of a finite collection of line segments in a plane.
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\end{proof}
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\subsubsection{\pending{Exercise 1.11.4d}}%
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\subsubsection{\verified{Exercise 1.11.4d}}%
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\hyperlabel{ssub:exercise-1.11.4d}
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$\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
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% FIXUP: padding
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\code{Bookshelf/Apostol/Chapter\_1\_11}
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\code*{Bookshelf/Apostol/Chapter\_1\_11}
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{Apostol.Chapter\_1\_11.exercise\_4d}
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\begin{proof}
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This is immediately proven by applying \nameref{sub:hermites-identity}.
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\end{proof}
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\subsubsection{\pending{Exercise 1.11.4e}}%
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\subsubsection{\verified{Exercise 1.11.4e}}%
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\hyperlabel{ssub:exercise-1.11.4e}
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$\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
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% FIXUP: padding
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\code{Bookshelf/Apostol/Chapter\_1\_11}
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\code*{Bookshelf/Apostol/Chapter\_1\_11}
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{Apostol.Chapter\_1\_11.exercise\_4e}
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\begin{proof}
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@ -1396,8 +1391,7 @@ The union of a finite collection of line segments in a plane.
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$\floor{nx}$.
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State and prove such a generalization.
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% FIXUP: padding
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\code{Bookshelf/Apostol/Chapter\_1\_11}
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\code*{Bookshelf/Apostol/Chapter\_1\_11}
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{Apostol.Chapter\_1\_11.exercise\_5}
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\begin{proof}
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@ -1626,8 +1620,7 @@ The union of a finite collection of line segments in a plane.
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$\sum_{n=1}^{b-1} \floor{na / b} = \sum_{n=1}^{b-1} \floor{a(b - n) / b}$.
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Now apply Exercises 4(a) and (b) to the bracket on the right.
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% FIXUP: padding
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\code{Bookshelf/Apostol/Chapter\_1\_11}
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\code*{Bookshelf/Apostol/Chapter\_1\_11}
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{Apostol.Chapter\_1\_11.exercise\_7b}
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\begin{proof}
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@ -88,10 +88,10 @@
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The \textbf{composition} of sets $F$ and $G$ is
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$$F \circ G = \{\pair{u, v} \mid \exists t(uGt \land tFv)\}.$$
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\lean{Mathlib/Data/Rel}{Rel.comp}
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\code{Bookshelf/Enderton/Set/Relation}{Set.Relation.comp}
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\lean{Mathlib/Data/Rel}{Rel.comp}
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\section{\defined{Connected}}%
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\hyperlabel{ref:connected}
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@ -106,10 +106,10 @@
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The \textbf{domain} of set $R$, denoted $\dom{R}$, is given by
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$$x \in \dom{R} \iff \exists y \pair{x, y} \in R.$$
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\lean{Mathlib/Data/Rel}{Rel.dom}
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\code{Bookshelf/Enderton/Set/Relation}{Set.Relation.dom}
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\lean{Mathlib/Data/Rel}{Rel.dom}
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\section{\defined{Empty Set Axiom}}%
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\hyperlabel{ref:empty-set-axiom}
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@ -202,10 +202,10 @@
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& = \{v \mid (\exists u \in A) uFv\}.
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\end{align*}
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\lean{Mathlib/Data/Rel}{Rel.image}
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\code{Bookshelf/Enderton/Set/Relation}{Set.Relation.image}
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\lean{Mathlib/Data/Rel}{Rel.image}
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\section{\defined{Inductive Set}}%
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\hyperlabel{ref:inductive-set}
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@ -247,10 +247,10 @@
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The \textbf{inverse} of a set $F$ is the set
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$$F^{-1} = \{\pair{u, v} \mid vFu\}.$$
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\lean{Mathlib/Data/Rel}{Rel.inv}
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\code{Bookshelf/Enderton/Set/Relation}{Set.Relation.inv}
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\lean{Mathlib/Data/Rel}{Rel.inv}
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\section{\defined{Irreflexive}}%
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\hyperlabel{ref:irreflexive}
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@ -322,9 +322,9 @@
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For any sets $u$ and $v$, the \textbf{ordered pair} $\pair{u, v}$ is
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the set $\{\{u\}, \{u, v\}\}$.
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\lean*{Prelude}{Prod}
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\code*{Bookshelf/Enderton/Set/OrderedPair}{OrderedPair}
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\code{Bookshelf/Enderton/Set/OrderedPair}{OrderedPair}
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\lean{Prelude}{Prod}
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\section{\defined{Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}}%
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\hyperlabel{ref:ordering-natural-numbers}
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@ -432,28 +432,28 @@
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The \textbf{range} of set $R$, denoted $\ran{R}$, is given by
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$$x \in \ran{R} \iff \exists t \pair{t, x} \in R.$$
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\lean{Mathlib/Data/Rel}{Rel.codom}
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\code{Bookshelf/Enderton/Set/Relation}{Set.Relation.ran}
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\lean{Mathlib/Data/Rel}{Rel.codom}
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\section{\defined{Reflexive}}%
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\hyperlabel{ref:reflexive}
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A binary relation $R$ is \textbf{reflexive} on $A$ if and only if $xRx$ for
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all $x \in A$.
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\lean*{Mathlib/Init/Algebra/Classes}{IsRefl}
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\code*{Bookshelf/Enderton/Set/Relation}{Set.Relation.isReflexive}
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\code{Bookshelf/Enderton/Set/Relation}{Set.Relation.isReflexive}
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\lean{Mathlib/Init/Algebra/Classes}{IsRefl}
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\section{\defined{Relation}}%
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\hyperlabel{ref:relation}
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A \textbf{relation} is a set of \nameref{ref:ordered-pair}s.
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\lean*{Mathlib/Data/Rel}{Rel}
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\code*{Bookshelf/Enderton/Set/Relation}{Set.Relation}
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\code{Bookshelf/Enderton/Set/Relation}{Set.Relation}
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\lean{Mathlib/Data/Rel}{Rel}
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\section{\defined{Restriction}}%
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\hyperlabel{ref:restriction}
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@ -507,9 +507,9 @@
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A binary relation $R$ is \textbf{transitive} if and only if whenever $xRy$ and
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$yRz$, then $xRz$.
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\lean*{Mathlib/Init/Algebra/Classes}{IsTrans}
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\code*{Bookshelf/Enderton/Set/Relation}{Set.Relation.isTransitive}
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\code{Bookshelf/Enderton/Set/Relation}{Set.Relation.isTransitive}
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\lean{Mathlib/Init/Algebra/Classes}{IsTrans}
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\section{\defined{Transitive Set}}%
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\hyperlabel{ref:transitive-set}
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@ -887,29 +887,24 @@
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\hyperlabel{sub:commutative-laws}
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For any sets $A$ and $B$,
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\begin{align*}
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A \cup B = B \cup A \\
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A \cap B = B \cap A
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\end{align*}
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\lean{Mathlib/Data/Set/Basic}{Set.union\_comm}
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\begin{enumerate}[(i)]
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\item $A \cup B = B \cup A$
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\item $A \cap B = B \cap A$
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\end{enumerate}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.commutative\_law\_i}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_comm}
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\lean{Mathlib/Data/Set/Basic}{Set.union\_comm}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.commutative\_law\_ii}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_comm}
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\begin{proof}
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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 $A \cup B = B \cup A$
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\item $A \cap B = B \cap A$.
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\end{enumerate}
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\paragraph{(i)}%
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@ -937,29 +932,24 @@
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\hyperlabel{sub:associative-laws}
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For any sets $A$, $B$ and $C$,
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\begin{align*}
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A \cup (B \cup C) & = (A \cup B) \cup C \\
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A \cap (B \cap C) & = (A \cap B) \cap C
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\end{align*}
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\lean{Mathlib/Data/Set/Basic}{Set.union\_assoc}
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\begin{enumerate}[(i)]
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\item $A \cup (B \cup C) = (A \cup B) \cup C$
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\item $A \cap (B \cap C) = (A \cap B) \cap C$
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\end{enumerate}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.associative\_law\_i}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_assoc}
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\lean{Mathlib/Data/Set/Basic}{Set.union\_assoc}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.associative\_law\_ii}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_assoc}
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\begin{proof}
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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 $A \cup (B \cup C) = (A \cup B) \cup C$
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\item $A \cap (B \cap C) = (A \cap B) \cap C$
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\end{enumerate}
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\paragraph{(i)}%
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@ -991,29 +981,24 @@
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\hyperlabel{sub:distributive-laws}
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For any sets $A$, $B$, and $C$,
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\begin{align*}
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A \cap (B \cup C) & = (A \cap B) \cup (A \cap C) \\
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A \cup (B \cap C) & = (A \cup B) \cap (A \cup C)
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\end{align*}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_distrib\_left}
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\begin{enumerate}[(i)]
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\item $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
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\item $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
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\end{enumerate}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.distributive\_law\_i}
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\lean{Mathlib/Data/Set/Basic}{Set.union\_distrib\_left}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_distrib\_left}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.distributive\_law\_ii}
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\lean{Mathlib/Data/Set/Basic}{Set.union\_distrib\_left}
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\begin{proof}
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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 $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
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\item $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
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\end{enumerate}
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\paragraph{(i)}%
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\item $C - (A \cap B) = (C - A) \cup (C - B)$
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\end{enumerate}
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\lean{Mathlib/Data/Set/Basic}{Set.diff\_inter\_diff}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.de\_morgans\_law\_i}
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\lean{Mathlib/Data/Set/Basic}{Set.diff\_inter}
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\lean{Mathlib/Data/Set/Basic}{Set.diff\_inter\_diff}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.de\_morgans\_law\_ii}
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\lean{Mathlib/Data/Set/Basic}{Set.diff\_inter}
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\begin{proof}
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Let $A$, $B$, and $C$ be sets.
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\item $A \cap (C - A) = \emptyset$
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\end{enumerate}
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\lean{Mathlib/Data/Set/Basic}{Set.union\_empty}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.emptyset\_identity\_i}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_empty}
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\lean{Mathlib/Data/Set/Basic}{Set.union\_empty}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.emptyset\_identity\_ii}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_diff\_self}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_empty}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.emptyset\_identity\_iii}
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_diff\_self}
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\begin{proof}
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Let $A$ be an arbitrary set.
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\hyperlabel{sub:monotonicity}
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For any sets $A$, $B$, and $C$,
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\begin{align*}
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A \subseteq B & \Rightarrow A \cup C \subseteq B \cup C \\
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A \subseteq B & \Rightarrow A \cap C \subseteq B \cap C \\
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A \subseteq B & \Rightarrow \bigcup A \subseteq \bigcup B
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\end{align*}
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\lean{Mathlib/Data/Set/Basic}
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{Set.union\_subset\_union\_left}
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\begin{enumerate}[(i)]
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\item $A \subseteq B \Rightarrow A \cup C \subseteq B \cup C$
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\item $A \subseteq B \Rightarrow A \cap C \subseteq B \cap C$
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\item $A \subseteq B \Rightarrow \bigcup A \subseteq \bigcup B$
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\end{enumerate}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.monotonicity\_i}
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\lean{Mathlib/Data/Set/Basic}
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{Set.inter\_subset\_inter\_left}
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{Set.union\_subset\_union\_left}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.monotonicity\_ii}
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\lean{Mathlib/Data/Set/Lattice}
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{Set.sUnion\_mono}
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\lean{Mathlib/Data/Set/Basic}
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{Set.inter\_subset\_inter\_left}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.monotonicity\_iii}
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\lean{Mathlib/Data/Set/Lattice}
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{Set.sUnion\_mono}
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\begin{proof}
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Let $A$, $B$, and $C$ be arbitrary sets.
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We prove that
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\begin{enumerate}[(i)]
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\item $A \subseteq B \Rightarrow A \cup C \subseteq B \cup C$
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\item $A \subseteq B \Rightarrow A \cap C \subseteq B \cap C$
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\item $A \subseteq B \Rightarrow \bigcup A \subseteq \bigcup B$
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\end{enumerate}
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\paragraph{(i)}%
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@ -1252,32 +1231,27 @@
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\hyperlabel{sub:anti-monotonicity}
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For any sets $A$, $B$, and $C$,
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\begin{align*}
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A \subseteq B & \Rightarrow C - B \subseteq C - A \\
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\emptyset \neq A \subseteq B & \Rightarrow \bigcap B \subseteq \bigcap A.
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\end{align*}
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\begin{enumerate}[(i)]
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\item $A \subseteq B \Rightarrow C - B \subseteq C - A$
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\item $\emptyset \neq A \subseteq B \Rightarrow
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\bigcap B \subseteq \bigcap A$.
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\end{enumerate}
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\code{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.anti\_monotonicity\_i}
|
||||
|
||||
\lean{Mathlib/Data/Set/Basic}
|
||||
{Set.diff\_subset\_diff\_right}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_2}
|
||||
{Enderton.Set.Chapter\_2.anti\_monotonicity\_i}
|
||||
{Enderton.Set.Chapter\_2.anti\_monotonicity\_ii}
|
||||
|
||||
\lean{Mathlib/Data/Set/Lattice}
|
||||
{Set.sInter\_subset\_sInter}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_2}
|
||||
{Enderton.Set.Chapter\_2.anti\_monotonicity\_ii}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $A$, $B$, and $C$ be arbitrary sets.
|
||||
We prove that
|
||||
\begin{enumerate}[(i)]
|
||||
\item $A \subseteq B \Rightarrow C - B \subseteq C - A$
|
||||
\item $\emptyset \neq A \subseteq B \Rightarrow
|
||||
\bigcap B \subseteq \bigcap A$
|
||||
\end{enumerate}
|
||||
|
||||
\paragraph{(i)}%
|
||||
|
||||
|
|
@ -1308,25 +1282,17 @@
|
|||
\hyperlabel{sub:general-distributive-laws}
|
||||
|
||||
For any sets $A$ and $\mathscr{B}$,
|
||||
\begin{align*}
|
||||
A \cup \bigcap \mathscr{B} & =
|
||||
\begin{enumerate}[(i)]
|
||||
\item $A \cup \bigcap \mathscr{B} =
|
||||
\bigcap\; \{ A \cup X \mid X \in \mathscr{B} \}
|
||||
\quad\text{for}\quad \mathscr{B} \neq \emptyset \\
|
||||
A \cap \bigcup \mathscr{B} & =
|
||||
\bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}
|
||||
\end{align*}
|
||||
\quad\text{for}\quad \mathscr{B} \neq \emptyset$
|
||||
\item $A \cap \bigcup \mathscr{B} =
|
||||
\bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $A$ and $\mathscr{B}$ be sets.
|
||||
We prove that
|
||||
\begin{enumerate}[(i)]
|
||||
\item For $\mathscr{B} \neq \emptyset$,
|
||||
$A \cup \bigcap \mathscr{B} =
|
||||
\bigcap\; \{ A \cup X \mid X \in \mathscr{B} \}$.
|
||||
\item $A \cap \bigcup \mathscr{B} =
|
||||
\bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}$
|
||||
\end{enumerate}
|
||||
|
||||
\paragraph{(i)}%
|
||||
|
||||
|
|
@ -1368,21 +1334,16 @@
|
|||
\hyperlabel{sub:general-de-morgans-laws}
|
||||
|
||||
For any set $C$ and $\mathscr{A} \neq \emptyset$,
|
||||
\begin{align*}
|
||||
C - \bigcup \mathscr{A} & = \bigcap\; \{ C - X \mid X \in \mathscr{A} \} \\
|
||||
C - \bigcap \mathscr{A} & = \bigcup\; \{ C - X \mid X \in \mathscr{A} \}
|
||||
\end{align*}
|
||||
\begin{enumerate}[(i)]
|
||||
\item $C - \bigcup \mathscr{A} =
|
||||
\bigcap\; \{ C - X \mid X \in \mathscr{A} \}$
|
||||
\item $C - \bigcap \mathscr{A} =
|
||||
\bigcup\; \{ C - X \mid X \in \mathscr{A} \}$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $C$ and $\mathscr{A}$ be sets such that $\mathscr{A} \neq \emptyset$.
|
||||
We prove that
|
||||
\begin{enumerate}[(i)]
|
||||
\item $C - \bigcup \mathscr{A} =
|
||||
\bigcap\; \{ C - X \mid X \in \mathscr{A} \}$
|
||||
\item $C - \bigcap \mathscr{A} =
|
||||
\bigcup\; \{ C - X \mid X \in \mathscr{A} \}$
|
||||
\end{enumerate}
|
||||
|
||||
\paragraph{(i)}%
|
||||
|
||||
|
|
@ -1431,12 +1392,12 @@
|
|||
Let $A$, $B$, and $C$ be sets.
|
||||
Then $A \cap (B - C) = (A \cap B) - C$.
|
||||
|
||||
\lean*{Mathlib/Data/Set/Basic}
|
||||
{Set.inter\_diff\_assoc}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_2}
|
||||
\code*{Bookshelf/Enderton/Set/Chapter\_2}
|
||||
{Enderton.Set.Chapter\_2.inter\_diff\_assoc}
|
||||
|
||||
\lean{Mathlib/Data/Set/Basic}
|
||||
{Set.inter\_diff\_assoc}
|
||||
|
||||
\begin{proof}
|
||||
Let $A$, $B$, and $C$ be sets.
|
||||
By definition of the intersection and relative complement of sets,
|
||||
|
|
@ -1953,12 +1914,12 @@
|
|||
|
||||
Show that $A \cap (B + C) = (A \cap B) + (A \cap C)$.
|
||||
|
||||
\lean*{Mathlib/Data/Set/Basic}
|
||||
{Set.inter\_symmDiff\_distrib\_left}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_2}
|
||||
\code*{Bookshelf/Enderton/Set/Chapter\_2}
|
||||
{Enderton.Set.Chapter\_2.exercise\_2\_15a}
|
||||
|
||||
\lean{Mathlib/Data/Set/Basic}
|
||||
{Set.inter\_symmDiff\_distrib\_left}
|
||||
|
||||
\begin{proof}
|
||||
By definition of the intersection, \nameref{ref:symmetric-difference}, and
|
||||
relative complement of sets,
|
||||
|
|
@ -1992,11 +1953,11 @@
|
|||
|
||||
Show that $A + (B + C) = (A + B) + C$.
|
||||
|
||||
\lean*{Mathlib/Order/SymmDiff}{symmDiff\_assoc}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_2}
|
||||
\code*{Bookshelf/Enderton/Set/Chapter\_2}
|
||||
{Enderton.Set.Chapter\_2.exercise\_2\_15b}
|
||||
|
||||
\lean{Mathlib/Order/SymmDiff}{symmDiff\_assoc}
|
||||
|
||||
\begin{proof}
|
||||
Let $A$, $B$, and $C$ be sets.
|
||||
We prove that
|
||||
|
|
@ -6094,13 +6055,13 @@
|
|||
\end{align}
|
||||
\end{theorem}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4i}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.add\_zero}
|
||||
|
||||
\lean{Init/Prelude}{Nat.add}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4i}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\paragraph{\eqref{sub:theorem-4i-eq1}}%
|
||||
|
|
@ -6129,13 +6090,13 @@
|
|||
\end{align}
|
||||
\end{theorem}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4j}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.mul\_zero}
|
||||
|
||||
\lean{Init/Prelude}{Nat.mul}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4j}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\paragraph{\eqref{sub:theorem-4j-eq1}}%
|
||||
|
|
@ -6159,11 +6120,11 @@
|
|||
In other words, $$0 + n = n.$$
|
||||
\end{lemma}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.zero\_add}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.left\_additive\_identity}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.zero\_add}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $S = \{n \in \omega \mid 0 + n = n\}$.
|
||||
|
|
@ -6203,11 +6164,11 @@
|
|||
In other words, $$m^+ + n = m + n^+.$$
|
||||
\end{lemma}
|
||||
|
||||
\lean{Std/Data/Nat/Lemmas}{Nat.succ\_add\_eq\_succ\_add}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.lemma\_2}
|
||||
|
||||
\lean{Std/Data/Nat/Lemmas}{Nat.succ\_add\_eq\_succ\_add}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $m \in \omega$ and define
|
||||
|
|
@ -6247,11 +6208,11 @@
|
|||
For $m, n, p \in \omega$, $$m + (n + p) = (m + n) + p.$$
|
||||
\end{theorem}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.add\_assoc}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4k\_1}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.add\_assoc}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Fix $n, p \in \omega$ and define
|
||||
|
|
@ -6304,11 +6265,11 @@
|
|||
For $m, n \in \omega$, $$m + n = n + m.$$
|
||||
\end{theorem}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.add\_comm}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4k\_2}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.add\_comm}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Fix $n \in \omega$ and define
|
||||
|
|
@ -6357,11 +6318,11 @@
|
|||
In other words, $$0 \cdot n = 0.$$
|
||||
\end{lemma}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.zero\_mul}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.zero\_multiplicand}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.zero\_mul}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Define
|
||||
|
|
@ -6409,11 +6370,11 @@
|
|||
In other words, $$m^+ \cdot n = m \cdot n + n.$$
|
||||
\end{lemma}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.succ\_mul}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.succ\_distrib}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.succ\_mul}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $m \in \omega$ and define
|
||||
|
|
@ -6466,11 +6427,11 @@
|
|||
For $m, n, p \in \omega$, $$m \cdot (n + p) = m \cdot n + m \cdot p.$$
|
||||
\end{theorem}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.left\_distrib}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4k\_3}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.left\_distrib}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Fix $n, p \in \omega$ and define
|
||||
|
|
@ -6529,11 +6490,11 @@
|
|||
In other words, $$m + 1 = m^+.$$
|
||||
\end{lemma}
|
||||
|
||||
\lean{Std/Data/Nat/Lemmas}{Nat.succ\_eq\_one\_add}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.succ\_identity}
|
||||
|
||||
\lean{Std/Data/Nat/Lemmas}{Nat.succ\_eq\_one\_add}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let
|
||||
|
|
@ -6583,11 +6544,11 @@
|
|||
In other words, $$m \cdot 1 = m.$$
|
||||
\end{lemma}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.mul\_one}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.right\_mul\_id}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.mul\_one}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let
|
||||
|
|
@ -6640,11 +6601,11 @@
|
|||
two properties in the opposite direction.
|
||||
\end{note}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.mul\_comm}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4k\_5}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.mul\_comm}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Fix $n \in \omega$ and define
|
||||
|
|
@ -6696,11 +6657,11 @@
|
|||
For $m, n, p \in \omega$, $$m \cdot (n \cdot p) = (m \cdot n) \cdot p.$$
|
||||
\end{theorem}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.mul\_assoc}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4k\_4}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.mul\_assoc}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Fix $m, n \in \omega$ and define
|
||||
|
|
@ -6858,11 +6819,11 @@
|
|||
No natural number is a member of itself.
|
||||
\end{lemma}
|
||||
|
||||
\lean{Init/Prelude}{Nat.lt\_irrefl}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
{Enderton.Set.Chapter\_4.theorem\_4l\_b}
|
||||
|
||||
\lean{Init/Prelude}{Nat.lt\_irrefl}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Define
|
||||
|
|
@ -7255,10 +7216,10 @@
|
|||
\end{align}
|
||||
\end{corollary}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.add\_right\_cancel}
|
||||
|
||||
\code{Common/Nat/Basic}{Nat.mul\_right\_cancel}
|
||||
|
||||
\lean{Init/Data/Nat/Basic}{Nat.add\_right\_cancel}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\paragraph{\eqref{sub:corollary-4p-eq1}}%
|
||||
|
|
|
|||
Loading…
Reference in New Issue