Simplify lean link formatting.
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@ -19,8 +19,8 @@
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\hyperlabel{ref:construction-sequence}
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A \textbf{construction sequence} is a finite sequence
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$\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that for
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each $i \leq n$ we have at least one of
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$\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that
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for each $i \leq n$ we have at least one of
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\begin{align*}
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& \epsilon_i \text{ is a sentence symbol} \\
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& \epsilon_i = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < i \\
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@ -39,8 +39,9 @@ An \textbf{expression} is a finite sequence of symbols.
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\hyperlabel{ref:well-formed-formula}
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An \nameref{ref:expression} that can be built up from the sentence symbols by
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applying some finite number of times the \textbf{formula-building operations}
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(on expressions) defined by the equations:
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applying some finite number of times the
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\textbf{formula-building operations} (on expressions) defined by the
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equations:
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\begin{align*}
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\mathcal{E}_{\neg}(\alpha)
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& = (\neg \alpha) \\
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@ -66,17 +67,13 @@ An \nameref{ref:expression} that can be built up from the sentence symbols by
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\hyperlabel{sec:lemma-0a}
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\begin{lemma}[0A]
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Assume that $\langle x_1, \ldots, x_m \rangle =
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\langle y_1, \ldots, y_m, \ldots, y_{m+k} \rangle$.
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Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
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\end{lemma}
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\begin{proof}
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TODO
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\end{proof}
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\chapter{Sentential Logic}%
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@ -89,16 +86,13 @@ Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
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\hyperlabel{sub:induction-principle-1}
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\begin{theorem}
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If $S$ is a set of wffs containing all the sentence symbols and closed under all
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five formula-building operations, then $S$ is the set of \textit{all} wffs.
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If $S$ is a set of wffs containing all the sentence symbols and closed under
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all five formula-building operations, then $S$ is the set of \textit{all}
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wffs.
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\end{theorem}
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\begin{proof}
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TODO
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\end{proof}
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\section{Exercises 1}%
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@ -113,9 +107,7 @@ The sentences shoudl be chosen so as to have an interesting structure, and the
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translations should each contain 15 or more symbols.
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\begin{answer}
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TODO
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\end{answer}
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\subsection{\sorry{Exercise 1.1.2}}%
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@ -124,11 +116,9 @@ The sentences shoudl be chosen so as to have an interesting structure, and the
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Show that there are no wffs of length 2, 3, or 6, but that any other positive
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length is possible.
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\begin{answer}
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\begin{proof}
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TODO
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\end{answer}
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\end{proof}
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\subsection{\sorry{Exercise 1.1.3}}%
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\hyperlabel{sub:exercise-1.1.3}
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@ -137,29 +127,26 @@ Let $\alpha$ be a wff; let $c$ be the number of places at which binary
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connective symbols $(\land, \lor, \Rightarrow, \Leftrightarrow)$ occur in
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$\alpha$; let $s$ be the number of places at which sentence symbols occur in
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$\alpha$.
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(For exmaple, if $\alpha$ is $(A \Rightarrow (\neg A))$ then $c = 1$ and $s = 2$.)
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(For example, if $\alpha$ is $(A \Rightarrow (\neg A))$ then $c = 1$ and
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$s = 2$.)
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Show by using the induction principle that $s = c + 1$.
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\begin{answer}
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\begin{proof}
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TODO
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\end{answer}
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\end{proof}
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\subsection{\sorry{Exercise 1.1.4}}%
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\hyperlabel{sub:exercise-1.1.4}
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Assume we have a construction sequence ending in $\phi$, where $\phi$ does not
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contain the symbol $A_4$.
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Suppose we delete all the expressions in the construction sequence that contain
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$A_4$.
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Suppose we delete all the expressions in the construction sequence that
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contain $A_4$.
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Show that the result is still a legal construction sequence.
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\begin{answer}
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\begin{proof}
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TODO
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\end{answer}
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\end{proof}
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\subsection{\sorry{Exercise 1.1.5}}%
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\hyperlabel{sub:exercise-1.1.5}
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@ -173,11 +160,9 @@ Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
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\textit{Suggestion}: Apply induction to show that the length is of the form
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$4k + 1$ and the number of sentence symbols is $k + 1$.
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\begin{answer}
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\begin{proof}
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TODO
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\end{answer}
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\end{proof}
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\subsection{\sorry{Exercise 1.1.6}}%
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\hyperlabel{sub:exercise-1.1.6}
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@ -189,10 +174,8 @@ Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
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\item Show that more than a quarter of the symbols are sentence symbols.
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\end{enumerate}
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\begin{answer}
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\begin{proof}
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TODO
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\end{answer}
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\end{proof}
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\end{document}
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File diff suppressed because it is too large
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@ -38,6 +38,153 @@ theorem commutative_law_ii (A B : Set α)
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exact and_comm
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_ = B ∩ A := rfl
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/-! #### Associative Laws
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For any sets `A`, `B`, and `C`,
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```
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A ∪ (B ∪ C) = (A ∪ B) ∪ C
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A ∩ (B ∩ C) = (A ∩ B) ∩ C
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```
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-/
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#check Set.union_assoc
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theorem associative_law_i (A B C : Set α)
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: A ∪ (B ∪ C) = (A ∪ B) ∪ C := calc A ∪ (B ∪ C)
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_ = { x | x ∈ A ∨ x ∈ B ∪ C } := rfl
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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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simp only [Set.mem_setOf_eq]
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rw [← or_assoc]
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_ = { x | x ∈ A ∪ B ∨ x ∈ C } := rfl
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_ = (A ∪ B) ∪ C := rfl
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#check Set.inter_assoc
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theorem associative_law_ii (A B C : Set α)
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: A ∩ (B ∩ C) = (A ∩ B) ∩ C := calc A ∩ (B ∩ C)
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_ = { x | x ∈ A ∧ (x ∈ B ∩ C) } := rfl
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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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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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/-! #### Distributive Laws
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For any sets `A`, `B`, and `C`,
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```
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A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
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A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
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```
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-/
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#check Set.inter_distrib_left
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theorem distributive_law_i (A B C : Set α)
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: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) := calc A ∩ (B ∪ C)
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_ = { x | x ∈ A ∧ x ∈ B ∪ C } := rfl
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_ = { x | x ∈ A ∧ (x ∈ B ∨ x ∈ C) } := rfl
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_ = { x | (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C) } := by
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ext _
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exact and_or_left
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_ = { x | x ∈ A ∩ B ∨ x ∈ A ∩ C } := rfl
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_ = (A ∩ B) ∪ (A ∩ C) := rfl
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#check Set.union_distrib_left
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theorem distributive_law_ii (A B C : Set α)
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: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) := calc A ∪ (B ∩ C)
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_ = { x | x ∈ A ∨ x ∈ B ∩ C } := rfl
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_ = { x | x ∈ A ∨ (x ∈ B ∧ x ∈ C) } := rfl
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_ = { x | (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C) } := by
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ext _
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exact or_and_left
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_ = { x | x ∈ A ∪ B ∧ x ∈ A ∪ C } := rfl
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_ = (A ∪ B) ∩ (A ∪ C) := rfl
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/-! #### De Morgan's Laws
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For any sets `A`, `B`, and `C`,
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```
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C - (A ∪ B) = (C - A) ∩ (C - B)
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C - (A ∩ B) = (C - A) ∪ (C - B)
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```
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-/
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#check Set.diff_inter_diff
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theorem de_morgans_law_i (A B C : Set α)
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: C \ (A ∪ B) = (C \ A) ∩ (C \ B) := calc C \ (A ∪ B)
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_ = { x | x ∈ C ∧ x ∉ A ∪ B } := rfl
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_ = { x | x ∈ C ∧ ¬(x ∈ A ∨ x ∈ B) } := rfl
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_ = { x | x ∈ C ∧ (x ∉ A ∧ x ∉ B) } := by
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ext _
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simp only [Set.mem_setOf_eq]
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rw [not_or_de_morgan]
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_ = { x | (x ∈ C ∧ x ∉ A) ∧ (x ∈ C ∧ x ∉ B) } := by
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ext _
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exact and_and_left
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_ = { x | x ∈ C \ A ∧ x ∈ C \ B } := rfl
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_ = (C \ A) ∩ (C \ B) := rfl
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#check Set.diff_inter
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theorem de_morgans_law_ii (A B C : Set α)
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: C \ (A ∩ B) = (C \ A) ∪ (C \ B) := calc C \ (A ∩ B)
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_ = { x | x ∈ C ∧ x ∉ A ∩ B } := rfl
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_ = { x | x ∈ C ∧ ¬(x ∈ A ∧ x ∈ B) } := rfl
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_ = { x | x ∈ C ∧ (x ∉ A ∨ x ∉ B) } := by
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ext _
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simp only [Set.mem_setOf_eq]
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rw [not_and_de_morgan]
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_ = { x | (x ∈ C ∧ x ∉ A) ∨ (x ∈ C ∧ x ∉ B) } := by
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ext _
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exact and_or_left
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_ = { x | x ∈ C \ A ∨ x ∈ C \ B } := rfl
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_ = (C \ A) ∪ (C \ B) := rfl
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/-! #### Identities Involving ∅
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For any set `A`,
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```
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A ∪ ∅ = A
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A ∩ ∅ = ∅
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A ∩ (C - A) = ∅
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```
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-/
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#check Set.union_empty
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theorem emptyset_identity_i (A : Set α)
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: A ∪ ∅ = A := calc A ∪ ∅
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_ = { x | x ∈ A ∨ x ∈ ∅ } := rfl
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_ = { x | x ∈ A ∨ False } := rfl
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_ = { x | x ∈ A } := by simp
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_ = A := rfl
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#check Set.inter_empty
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theorem emptyset_identity_ii (A : Set α)
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: A ∩ ∅ = ∅ := calc A ∩ ∅
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_ = { x | x ∈ A ∧ x ∈ ∅ } := rfl
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_ = { x | x ∈ A ∧ False } := rfl
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_ = { x | False } := by simp
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_ = ∅ := rfl
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#check Set.inter_diff_self
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theorem emptyset_identity_iii (A C : Set α)
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: A ∩ (C \ A) = ∅ := calc A ∩ (C \ A)
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_ = { x | x ∈ A ∧ x ∈ C \ A } := rfl
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_ = { x | x ∈ A ∧ (x ∈ C ∧ x ∉ A) } := rfl
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_ = { x | x ∈ C ∧ False } := by simp
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_ = { x | False } := by simp
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_ = ∅ := rfl
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/-- #### Exercise 2.1
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Assume that `A` is the set of integers divisible by `4`. Similarly assume that
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@ -19,7 +19,7 @@ namespace Enderton.Set.Chapter_3
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If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`.
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-/
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theorem theorem_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
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theorem lemma_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
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: OrderedPair x y ∈ 𝒫 𝒫 C := by
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have hxs : {x} ⊆ C := Set.singleton_subset_iff.mpr hx
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have hxys : {x, y} ⊆ C := Set.mem_mem_imp_pair_subset hx hy
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@ -22,11 +22,11 @@ Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
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\sum_{i=0}^n a_i = \frac{(n + 1)(a_0 + a_n)}{2}.
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\end{equation}
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\begin{proof}
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\code{Common/Real/Sequence/Arithmetic}
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{Real.Arithmetic.sum\_recursive\_closed}
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\begin{proof}
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Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
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By definition, for all $k \in \mathbb{N}$,
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\begin{equation}
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@ -82,11 +82,11 @@ Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
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\sum_{i=0}^n a_i = \frac{a_0(1 - r^{n+1})}{1 - r}.
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\end{equation}
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\begin{proof}
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\code{Common/Real/Sequence/Geometric}
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{Real.Geometric.sum\_recursive\_closed}
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\begin{proof}
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Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
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By definition, for all $k \in \mathbb{N}$,
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\begin{equation}
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50
preamble.tex
50
preamble.tex
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@ -44,61 +44,47 @@
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\label{#1}%
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\hypertarget{#1}{}}
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% Denote whether we are working with a standard/Mathlib statement (lean) or a
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% custom one (code).
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% Denote if working with a predefined statement/theorem or a custom one.
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\newcommand\@leanlink[4]{%
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\textcolor{blue}{\raisebox{-4.5pt}{%
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\tikz{\draw (0, 0) node[yscale=-1,xscale=1] {\faFont};}}}%
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{-\;}\href{#1/#2.html\##3}{#4}}
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\textcolor{BlueViolet}{\raisebox{-4.5pt}{%
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\tikz{\draw (0, 0) node[yscale=-1,xscale=1] {\faFont};}}{-\;}}%
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\href{#1/#2.html\##3}{\color{BlueViolet}{#4}}}
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\newcommand\@codelink[4]{%
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\textcolor{blue}{\raisebox{-4.5pt}{%
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\tikz{\draw (0, 0) node[] {\faCodeBranch};}}}%
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{-\;}\href{#1/#2.html\##3}{#4}}
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\textcolor{MidnightBlue}{\raisebox{-4.5pt}{%
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\tikz{\draw (0, 0) node[xshift=8pt] {\faCodeBranch};}}{-\;}}%
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\href{#1/#2.html\##3}{\color{MidnightBlue}{#4}}}
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% Reference to an anchor of Lean documentation.
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% Reference to an anchor of generated Lean documentation.
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\newcommand\leanref[3]{%
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\@leanlink{#1}{#2}{#3}{#3}\vspace{10pt}}
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\WithSuffix\newcommand\leanref*[3]{%
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\@leanlink{#1}{#2}{#3}{#3}}
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\newcommand\coderef[3]{%
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\@codelink{#1}{#2}{#3}{#3}\vspace{10pt}}
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\WithSuffix\newcommand\coderef*[3]{%
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\@codelink{#1}{#2}{#3}{#3}}
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% Variants that allows customizing display text.
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\newcommand\leanpref[4]{%
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\@leanlink{#1}{#2}{#3}{#4}\vspace{10pt}}
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\WithSuffix\newcommand\leanpref*[4]{%
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\@leanlink{#1}{#2}{#3}{#4}}
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\newcommand\codepref[4]{%
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\@codelink{#1}{#2}{#3}{#4}\vspace{10pt}}
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\WithSuffix\newcommand\codepref*[4]{%
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\@codelink{#1}{#2}{#3}{#4}}
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% Macro to build all Lean related commands relative to a specified directory.
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\newcommand\makeleancommands[1]{%
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\newcommand\lean[2]{%
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\leanref{#1}{##1}{##2}}
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\noindent\leanref{#1}{##1}{##2}}
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\WithSuffix\newcommand\lean*[2]{%
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\leanref*{#1}{##1}{##2}}
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\vspace{6pt}\noindent\leanref{#1}{##1}{##2}}
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\newcommand\code[2]{%
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\coderef{#1}{##1}{##2}}
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\noindent\coderef{#1}{##1}{##2}}
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\WithSuffix\newcommand\code*[2]{%
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\coderef*{#1}{##1}{##2}}
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\vspace{6pt}\noindent\coderef{#1}{##1}{##2}}
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\newcommand\leanp[3]{%
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\leanpref{#1}{##1}{##2}{##3}}
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\noindent\leanpref{#1}{##1}{##2}{##3}}
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\WithSuffix\newcommand\leanp*[3]{%
|
||||
\leanpref*{#1}{##1}{##2}{##3}}
|
||||
\vspace{6pt}\noindent\leanpref{#1}{##1}{##2}{##3}}
|
||||
|
||||
\newcommand\codep[3]{%
|
||||
\codepref{#1}{##1}{##2}{##3}}
|
||||
\noindent\codepref{#1}{##1}{##2}{##3}}
|
||||
\WithSuffix\newcommand\codep*[3]{%
|
||||
\codepref*{#1}{##1}{##2}{##3}}
|
||||
\vspace{6pt}\noindent\codepref{#1}{##1}{##2}{##3}}
|
||||
}
|
||||
|
||||
% ========================================
|
||||
|
|
@ -124,11 +110,7 @@
|
|||
|
||||
\newcommand\@statement[1]{%
|
||||
\linedivider*\paragraph{\normalfont\normalsize\textit{#1.}}}
|
||||
\newcommand{\statementpadding}{\ \vspace{8pt}}
|
||||
|
||||
\newenvironment{answer}{\@statement{Answer}}{\hfill$\square$}
|
||||
\newenvironment{axiom}{\@statement{Axiom}}{\hfill$\square$}
|
||||
\newenvironment{definition}{\@statement{Definition}}{\hfill$\square$}
|
||||
\renewenvironment{proof}{\@statement{Proof}}{\hfill$\square$}
|
||||
|
||||
\newtheorem{corollaryinner}{Corollary}
|
||||
|
|
|
|||
Loading…
Reference in New Issue