Enderton. Reorder theorems in Lean modules and add set/logic prompts.
parent
2c1547aee9
commit
f990a7a30a
|
|
@ -8,26 +8,21 @@ namespace Apostol.Chapter_1_11
|
|||
|
||||
open BigOperators
|
||||
|
||||
/-! ## Exercise 4
|
||||
|
||||
Prove that the greatest-integer function has the properties indicated.
|
||||
-/
|
||||
|
||||
/-- ### Exercise 4a
|
||||
/-- #### Exercise 4a
|
||||
|
||||
`⌊x + n⌋ = ⌊x⌋ + n` for every integer `n`.
|
||||
-/
|
||||
theorem exercise_4a (x : ℝ) (n : ℤ) : ⌊x + n⌋ = ⌊x⌋ + n :=
|
||||
Int.floor_add_int x n
|
||||
|
||||
/-- ### Exercise 4b.1
|
||||
/-- #### Exercise 4b.1
|
||||
|
||||
`⌊-x⌋ = -⌊x⌋` if `x` is an integer.
|
||||
-/
|
||||
theorem exercise_4b_1 (x : ℤ) : ⌊-x⌋ = -⌊x⌋ := by
|
||||
simp only [Int.floor_int, id_eq]
|
||||
|
||||
/-- ### Exercise 4b.2
|
||||
/-- #### Exercise 4b.2
|
||||
|
||||
`⌊-x⌋ = -⌊x⌋ - 1` otherwise.
|
||||
-/
|
||||
|
|
@ -47,7 +42,7 @@ theorem exercise_4b_2 (x : ℝ) (h : ∃ n : ℤ, x ∈ Set.Ioo ↑n (↑n + (1
|
|||
· exact (Set.mem_Ioo.mp hn).left
|
||||
· exact le_of_lt (Set.mem_Ico.mp hn').right
|
||||
|
||||
/-- ### Exercise 4c
|
||||
/-- #### Exercise 4c
|
||||
|
||||
`⌊x + y⌋ = ⌊x⌋ + ⌊y⌋` or `⌊x⌋ + ⌊y⌋ + 1`.
|
||||
-/
|
||||
|
|
@ -77,7 +72,7 @@ theorem exercise_4c (x y : ℝ)
|
|||
rw [← sub_lt_iff_lt_add', ← sub_sub, add_sub_cancel, add_sub_cancel]
|
||||
exact add_lt_add (Int.fract_lt_one x) (Int.fract_lt_one y)
|
||||
|
||||
/-- ### Exercise 5
|
||||
/-- #### Exercise 5
|
||||
|
||||
The formulas in Exercises 4(d) and 4(e) suggest a generalization for `⌊nx⌋`.
|
||||
State and prove such a generalization.
|
||||
|
|
@ -86,7 +81,7 @@ theorem exercise_5 (n : ℕ) (x : ℝ)
|
|||
: ⌊n * x⌋ = Finset.sum (Finset.range n) (fun i => ⌊x + i/n⌋) :=
|
||||
Real.Floor.floor_mul_eq_sum_range_floor_add_index_div n x
|
||||
|
||||
/-- ### Exercise 4d
|
||||
/-- #### Exercise 4d
|
||||
|
||||
`⌊2x⌋ = ⌊x⌋ + ⌊x + 1/2⌋`
|
||||
-/
|
||||
|
|
@ -99,7 +94,7 @@ theorem exercise_4d (x : ℝ)
|
|||
simp
|
||||
rw [add_comm]
|
||||
|
||||
/-- ### Exercise 4e
|
||||
/-- #### Exercise 4e
|
||||
|
||||
`⌊3x⌋ = ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋`
|
||||
-/
|
||||
|
|
@ -113,7 +108,7 @@ theorem exercise_4e (x : ℝ)
|
|||
conv => rhs; rw [← add_rotate']; arg 2; rw [add_comm]
|
||||
rw [← add_assoc]
|
||||
|
||||
/-- ### Exercise 7b
|
||||
/-- #### Exercise 7b
|
||||
|
||||
If `a` and `b` are positive integers with no common factor, we have the formula
|
||||
`∑_{n=1}^{b-1} ⌊na / b⌋ = ((a - 1)(b - 1)) / 2`. When `b = 1`, the sum on the
|
||||
|
|
@ -130,7 +125,7 @@ theorem exercise_7b (ha : a > 0) (hb : b > 0) (hp : Nat.coprime a b)
|
|||
|
||||
section
|
||||
|
||||
/-- ### Exercise 8
|
||||
/-- #### Exercise 8
|
||||
|
||||
Let `S` be a set of points on the real line. The *characteristic function* of
|
||||
`S` is, by definition, the function `Χ` such that `Χₛ(x) = 1` for every `x` in
|
||||
|
|
|
|||
|
|
@ -489,8 +489,6 @@ theorem forall_mem_le_forall_mem_imp_sup_le_inf (S T : Set ℝ)
|
|||
_ < x := hx.right
|
||||
simp at this
|
||||
|
||||
/-! ## Exercises -/
|
||||
|
||||
/-- #### Exercise 1
|
||||
|
||||
If `x` and `y` are arbitrary real numbers with `x < y`, prove that there is at
|
||||
|
|
|
|||
|
|
@ -88,6 +88,71 @@
|
|||
\lean*{Init/Prelude}
|
||||
{Prod}
|
||||
|
||||
\section{\defined{Tautological Implication}}%
|
||||
\hyperlabel{ref:tautological-implication}
|
||||
|
||||
Consider a set $\Sigma$ of \nameref{ref:well-formed-formula}s and another wff
|
||||
$\tau$.
|
||||
$\Sigma$ \textbf{tautologically implies} $\tau$ (written $\Sigma \vDash \tau$)
|
||||
if and only if every \nameref{ref:truth-assignment} for the sentence symbols
|
||||
in $\Sigma$ and $\tau$ that satisfies every member of $\Sigma$ also
|
||||
satisfies $\tau$.
|
||||
If $\Sigma$ is singleton $\{\sigma\}$, then we write "$\sigma \vDash \tau$" in
|
||||
place of "$\{\sigma\} \vDash \tau$."
|
||||
|
||||
If both $\sigma \vDash \tau$ and $\tau \vDash \sigma$, then $\sigma$ and
|
||||
$\tau$ are said to be \textbf{tautologically equivalent} (written
|
||||
$\sigma \vDash \Dashv \tau$).
|
||||
|
||||
\section{\defined{Truth Assignment}}%
|
||||
\hyperlabel{ref:truth-assignment}
|
||||
|
||||
A \textbf{truth assignment} $v$ for a set $\mathcal{S}$ of sentence symbols is
|
||||
a function $$v \colon \mathcal{S} \rightarrow \{F, T\}$$ assigning either
|
||||
$T$ or $F$ to each symbol in $\mathcal{S}$.
|
||||
|
||||
Let $\bar{\mathcal{S}}$ be the set of \nameref{ref:well-formed-formula}s that
|
||||
can be built up from $\mathcal{S}$ by the five
|
||||
\nameref{ref:formula-building-operations}.
|
||||
We define \textbf{extension} $\bar{v}$ of $v$,
|
||||
$$\bar{v} \colon \bar{\mathcal{S}} \rightarrow \{F, T\},$$
|
||||
as the function that satisfies the following conditions for any
|
||||
$\alpha, \beta \in \mathcal{S}$:
|
||||
\begin{enumerate}[(1)]
|
||||
\setcounter{enumi}{-1}
|
||||
\item For any $A \in \mathcal{S}$, $\bar{v}(A) = v(A)$.
|
||||
(Thus $\bar{v}$ is indeed an extension of $v$.)
|
||||
\item $\bar{v}((\neg\alpha)) = \begin{cases}
|
||||
T & \text{if } \bar{v}(\alpha) = F, \\
|
||||
F & \text{otherwise}.
|
||||
\end{cases}$
|
||||
\item $\bar{v}((\alpha \land \beta)) = \begin{cases}
|
||||
T & \text{if } \bar{v}(\alpha) = T \text{ and } \bar{v}(\beta) = T, \\
|
||||
F & \text{otherwise}.
|
||||
\end{cases}$
|
||||
\item $\bar{v}((\alpha \lor \beta)) = \begin{cases}
|
||||
T & \text{if } \bar{v}(\alpha) = T \text{ or }
|
||||
\bar{v}(\beta) = T \text{ (or both)}, \\
|
||||
F & \text{otherwise}.
|
||||
\end{cases}$
|
||||
\item $\bar{v}((\alpha \Rightarrow \beta)) = \begin{cases}
|
||||
F & \text{if } \bar{v}(\alpha) = T \text{ and } \bar{v}(\beta) = F, \\
|
||||
T & \text{otherwise}.
|
||||
\end{cases}$
|
||||
\item $\bar{v}((\alpha \Leftrightarrow \beta)) = \begin{cases}
|
||||
T & \text{if } \bar{v}(\alpha) = \bar{v}(\beta), \\
|
||||
F & \text{otherwise}.
|
||||
\end{cases}$
|
||||
\end{enumerate}
|
||||
We say that truth assignment $v$ \textbf{satisfies} $\phi$ if and only if
|
||||
$\bar{v}(\phi) = T$.
|
||||
|
||||
\lean*{Init/Prelude}
|
||||
{True}
|
||||
|
||||
\lean{Init/Prelude}
|
||||
{False}
|
||||
|
||||
\section{\defined{Well-Formed Formula}}%
|
||||
\hyperlabel{ref:well-formed-formula}
|
||||
|
||||
|
|
@ -164,16 +229,17 @@
|
|||
\hyperlabel{sub:induction-principle-1}
|
||||
|
||||
\begin{theorem}
|
||||
If $S$ is a set of wffs containing all the sentence symbols and closed under
|
||||
all five formula-building operations, then $S$ is the set of \textit{all}
|
||||
wffs.
|
||||
If $S$ is a set of \nameref{ref:well-formed-formula}s containing all the
|
||||
sentence symbols and closed under all five
|
||||
\nameref{ref:formula-building-operations}, then $S$ is the set of
|
||||
\textit{all} wffs.
|
||||
\end{theorem}
|
||||
|
||||
\code{Bookshelf/Enderton/Logic/Chapter\_1}
|
||||
{Enderton.Logic.Chapter\_1.Wff.rec}
|
||||
|
||||
\begin{proof}
|
||||
We note every \nameref{ref:well-formed-formula} can be characterized by a
|
||||
We note every well-formed formula can be characterized by a
|
||||
\nameref{ref:construction-sequence}.
|
||||
For natural number $m$, let $P(m)$ be the statement:
|
||||
\begin{induction}
|
||||
|
|
@ -251,7 +317,8 @@
|
|||
\hyperlabel{sub:balanced-parentheses}
|
||||
|
||||
\begin{lemma}
|
||||
All well-formed formulas have an equal number of left and right parentheses.
|
||||
All \nameref{ref:well-formed-formula}s have an equal number of left and
|
||||
right parentheses.
|
||||
\end{lemma}
|
||||
|
||||
\begin{proof}
|
||||
|
|
@ -300,8 +367,8 @@
|
|||
\hyperlabel{sub:parentheses-count}
|
||||
|
||||
\begin{lemma}
|
||||
Let $\phi$ be a well-formed formula and $c$ be the number of places at which
|
||||
a sentential connective symbol exists.
|
||||
Let $\phi$ be a \nameref{ref:well-formed-formula} and $c$ be the number of
|
||||
places at which a sentential connective symbol exists.
|
||||
Then there is $2c$ parentheses in $\phi$.
|
||||
\end{lemma}
|
||||
|
||||
|
|
@ -350,6 +417,36 @@
|
|||
|
||||
\end{proof}
|
||||
|
||||
\section{Truth Assignments}%
|
||||
\hyperlabel{sec:truth-assignments}
|
||||
|
||||
\subsection{\sorry{Theorem 12A}}%
|
||||
\hyperlabel{sub:theorem-12a}
|
||||
|
||||
\begin{theorem}[12A]
|
||||
For any \nameref{ref:truth-assignment} $v$ for a set $\mathcal{S}$ there is
|
||||
a unique extension
|
||||
$\bar{v} \colon \bar{\mathcal{S}} \rightarrow \{F, T\}$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Compactness Theorem}}%
|
||||
\hyperlabel{sub:compactness-theorem}
|
||||
|
||||
\begin{theorem}
|
||||
Let $\Sigma$ be an infinite set of \nameref{ref:well-formed-formula}s such
|
||||
that for any finite subset $\Sigma_0$ of $\Sigma$, there is a
|
||||
\nameref{ref:truth-assignment} that satisfies every member of $\Sigma_0$.
|
||||
Then there is a truth assignment that satisfies every member of $\Sigma$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\section{Exercises 1}%
|
||||
\hyperlabel{sec:exercises-1}
|
||||
|
||||
|
|
@ -699,4 +796,300 @@
|
|||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.1}}%
|
||||
\hyperlabel{sub:exercise-1.2.1}
|
||||
|
||||
Show that neither of the following two formulas tautologically implies the
|
||||
other:
|
||||
\begin{gather*}
|
||||
(A \Leftrightarrow (B \Leftrightarrow C)), \\
|
||||
((A \land (B \land C)) \lor ((\neg A) \land ((\neg B) \land (\neg C)))).
|
||||
\end{gather*}
|
||||
\textit{Suggestion}: Only two \nameref{ref:truth-assignment}s are needed, not
|
||||
eight.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.2a}}%
|
||||
\hyperlabel{sub:exercise-1.2.2a}
|
||||
|
||||
Is $(((P \Rightarrow Q) \Rightarrow P) \Rightarrow P)$ a tautology?
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.2b}}%
|
||||
\hyperlabel{sub:exercise-1.2.2b}
|
||||
|
||||
Define $\sigma_k$ recursively as follows: $\sigma_0 = (P \Rightarrow Q)$ and
|
||||
$\sigma_{k + 1} = (\sigma_k \Rightarrow P)$. For which values of $k$ is
|
||||
$\sigma_k$ a tautology? (Part (a) corresponds to $k = 2$.)
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.3a}}%
|
||||
\hyperlabel{sub:exercise-1.2.3a}
|
||||
|
||||
Determine whether or not $((P \Rightarrow Q) \lor (Q \Rightarrow P))$ is a
|
||||
tautology.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.3b}}%
|
||||
\hyperlabel{sub:exercise-1.2.3b}
|
||||
|
||||
Determine whether or not $((P \land Q) \Rightarrow R)$ tautologically implies
|
||||
$((P \Rightarrow R) \lor (Q \Rightarrow R))$.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.4}}%
|
||||
\hyperlabel{sub:exercise-1.2.4}
|
||||
|
||||
Show that the following hold:
|
||||
\begin{enumerate}[(a)]
|
||||
\item $\Sigma; \alpha \vDash \beta$ iff
|
||||
$\Sigma \vDash (\alpha \Rightarrow \beta)$.
|
||||
\item $\alpha \vDash \Dashv \beta$ iff
|
||||
$\vDash (\alpha \Leftrightarrow \beta)$.
|
||||
\end{enumerate}
|
||||
(Recall that $\Sigma; \alpha = \Sigma \cup \{\alpha\}$, the set $\Sigma$
|
||||
together with the one possibly new member $\alpha$.)
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.5}}%
|
||||
\hyperlabel{sub:exercise-1.2.5}
|
||||
|
||||
Prove or refute each of the following assertions:
|
||||
\begin{enumerate}[(a)]
|
||||
\item If either $\Sigma \vDash \alpha$ or $\Sigma \vDash \beta$, then
|
||||
$\Sigma \vDash (\alpha \lor \beta)$.
|
||||
\item If $\Sigma \vDash (\alpha \lor \beta)$, then either
|
||||
$\Sigma \vDash \alpha$ or $\Sigma \vDash \beta$.
|
||||
\end{enumerate}
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.6a}}%
|
||||
\hyperlabel{sub:exercise-1.2.6a}
|
||||
|
||||
Show that if $v_1$ and $v_2$ are \nameref{ref:truth-assignment}s which agree
|
||||
on all the sentence symbols in the wff $\alpha$, then
|
||||
$\bar{v}_1(\alpha) = \bar{v}_2(\alpha)$.
|
||||
Use the \nameref{sub:induction-principle-1}.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.6b}}%
|
||||
\hyperlabel{sub:exercise-1.2.6b}
|
||||
|
||||
Let $\mathcal{S}$ be a set of sentence symbols that includes those in $\Sigma$
|
||||
and $\tau$ (and possibly more).
|
||||
Show that $\Sigma \vDash \tau$ iff every truth assignment for $\mathcal{S}$
|
||||
which satisfies every member of $\Sigma$ also satisfies $\tau$.
|
||||
(This is an easy consequence of part (a). The point of part (b) is that we do
|
||||
not need to worry aboutgetting the domain of a truth assignment
|
||||
\textit{exactly} perfect, as long as it is big enough. For example, one
|
||||
option would be always to use truth assignments on the set of \textit{all}
|
||||
sentence symbols. The drawback is that these are infinite objects, and there
|
||||
are a great many -- uncountably many -- of them.)
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.7}}%
|
||||
\hyperlabel{sub:exercise-1.2.7}
|
||||
|
||||
You are in a land inhabited by people who either always tell the truth or
|
||||
always tell falsehoods.
|
||||
You come to a fork in the road and you need to know which fork leads to the
|
||||
capital.
|
||||
There is a local resident there, but he has time only to reply to one
|
||||
yes-or-no question.
|
||||
What one question should you ask so as to learn which fork to take?
|
||||
\textit{Suggestion}: Make a table.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.8}}%
|
||||
\hyperlabel{sub:exercise-1.2.8}
|
||||
|
||||
(Substitution) Consider a suquence $\alpha_1, \alpha_2, \ldots$ of wffs.
|
||||
For each wff $\phi$ let $\phi^*$ be the result of replacing the sentence
|
||||
symbol $A_n$ by $\alpha_n$ for each $n$.
|
||||
|
||||
\subsubsection{\sorry{Exercise 1.2.8a}}%
|
||||
\hyperlabel{ssub:exercise-1.2.8a}
|
||||
|
||||
Let $v$ be a truth assignment for the set of all sentence symbols; define $u$
|
||||
to be the truth assignment for which $u(A_n) = \bar{v}(\alpha_n)$.
|
||||
Show that $\bar{u}(\phi) = \bar{v}(\phi^*)$.
|
||||
Use the induction principle.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsubsection{\sorry{Exercise 1.2.8b}}%
|
||||
\hyperlabel{ssub:exercise-1.2.8b}
|
||||
|
||||
Show that if $\phi$ is a tautology, then so is $\phi^*$.
|
||||
(For example, one of our selected tautologies is
|
||||
$((A \land B) \Leftrightarrow (B \land A))$. From this we can conclude, by
|
||||
substitution, that
|
||||
$((\alpha \land \beta) \Leftrightarrow (\beta \land \alpha))$ is a
|
||||
tautology, for any wffs $\alpha$ and $\beta$.)
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.9}}%
|
||||
\hyperlabel{sub:exercise-1.2.9}
|
||||
|
||||
(Duality) Let $\alpha$ be a wff whose only connective symbols are $\land$,
|
||||
$\lor$, and $\neg$.
|
||||
Let $\alpha^*$ be the result of interchanging $\land$ and $\lor$ and replacing
|
||||
each sentence symbol by its negation.
|
||||
Show that $\alpha^*$ is tautologically equivalent to $(\neg \alpha)$.
|
||||
Use the \nameref{sub:induction-principle-1}.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.10}}%
|
||||
\hyperlabel{sub:exercise-1.2.10}
|
||||
|
||||
Say that a set $\Sigma_1$ of wffs is \textit{equivalent} to a set $\Sigma_2$
|
||||
of wffs iff for any wff $\alpha$, we have $\Sigma_1 \vDash \alpha$ iff
|
||||
$\Sigma_2 \vDash \alpha$.
|
||||
A set $\Sigma$ is \textit{independent} iff no member of $\Sigma$ is
|
||||
tautologically implied by the remaining members in $\Sigma$.
|
||||
Show that the following hold.
|
||||
|
||||
\subsubsection{\sorry{Exercise 1.2.10a}}%
|
||||
\hyperlabel{ssub:exercise-1.2.10a}
|
||||
|
||||
A finite set of wffs has an independent equivalent subset.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsubsection{\sorry{Exercise 1.2.10b}}%
|
||||
\hyperlabel{ssub:exercise-1.2.10b}
|
||||
|
||||
An infinite set need not have an independent equivalent subset.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsubsection{\sorry{Exercise 1.2.10c}}%
|
||||
\hyperlabel{ssub:exercise-1.2.10c}
|
||||
|
||||
Let $\Sigma = \{\sigma_0, \sigma_1, \ldots\}$; show that there is an
|
||||
independent equivalent set $\Sigma'$.
|
||||
(By part (b), we cannot hope to have $\Sigma' \subseteq \Sigma$ in general.)
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.11}}%
|
||||
\hyperlabel{sub:exercise-1.2.11}
|
||||
|
||||
Show that a truth assignment $v$ satisfies the wff
|
||||
$$(\cdots (A_1 \Leftrightarrow A_2)
|
||||
\Leftrightarrow \cdots \Leftrightarrow A_n)$$
|
||||
iff $v(A_i) = F$ for an even number of $i$'s, $1 \leq i \leq n$.
|
||||
(By the associative law for $\Leftrightarrow$, the placement of the
|
||||
parentheses is not crucial.)
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.12}}%
|
||||
\hyperlabel{sub:exercise-1.2.12}
|
||||
|
||||
There are three suspects for a murder: Adams, Brown, and Clark.
|
||||
Adams says "I didn't do it. The victim was an old acquaintance of Brown's.
|
||||
But Clark hated him."
|
||||
Brown states "I didn't do it. I didn't even know the guy. Besides I was out of
|
||||
town all that week."
|
||||
Clark says "I didn't do it. I saw both ADams and Brown downtown with the
|
||||
victim that day; one of them must have done it."
|
||||
Assume that the two innocent men are telling the truth, but that the guilty
|
||||
man might not be.
|
||||
Who did it?
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.13}}%
|
||||
\hyperlabel{sub:exercise-1.2.13}
|
||||
|
||||
An advertisement for a tennis magazine states, "If I'm not playing tennis,
|
||||
I'm watching tennis. And if I'm not watching tennis, I'm reading about
|
||||
tennis."
|
||||
We can assume that the speaker cannot do more than one of these activities at
|
||||
a time.
|
||||
What is the speaker doing?
|
||||
(Translate the given sentences into our formal language; consider the possible
|
||||
truth assignments.)
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.14}}%
|
||||
\hyperlabel{sub:exercise-1.2.14}
|
||||
|
||||
Let $\mathcal{S}$ be the set of all sentence symbols, and assume that
|
||||
$v \colon \mathcal{S} \rightarrow \{F, T\}$ is a truth assignment.
|
||||
Show there is \textit{at most} one extension $\bar{v}$ meeting conditions 0-5
|
||||
listed at the beginning of this section.
|
||||
(Suppose that $\bar{v}_1$ and $\bar{v}_2$ are both such extensions. Use the
|
||||
\nameref{sub:induction-principle-1} to show that $\bar{v}_1 = \bar{v}_2$.
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 1.2.15}}%
|
||||
\hyperlabel{sub:exercise-1.2.15}
|
||||
|
||||
Of the following three formulas, which tautologically implies which?
|
||||
\begin{enumerate}[(a)]
|
||||
\item $(A \Leftrightarrow B)$
|
||||
\item $(\neg((A \Rightarrow B) \Rightarrow (\neg(B \Rightarrow A))))$
|
||||
\item $(((\neg A) \lor B) \land (A \lor (\neg B)))$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{proof}
|
||||
TODO
|
||||
\end{proof}
|
||||
|
||||
\end{document}
|
||||
|
|
|
|||
|
|
@ -213,11 +213,9 @@
|
|||
\textbf{maps $A$ into $B$} (written $F \colon A \rightarrow B$) iff $F$ is a
|
||||
function, $\dom{F} = A$, and $\ran{F} \subseteq B$.
|
||||
If $\ran{F} = B$, then $F$ is a function from \textbf{$A$ onto $B$}.
|
||||
|
||||
A function $F$ is \textbf{one-to-one} iff for each $y \in \ran{F}$ there is
|
||||
only one $x$ such that $xFy$.
|
||||
One-to-one functions are sometimes called \textbf{injections}.
|
||||
|
||||
A one-to-one function from $A$ onto $B$ is a
|
||||
\textbf{one-to-one correspondence} between $A$ and $B$.
|
||||
|
||||
|
|
@ -6173,7 +6171,7 @@
|
|||
\hyperlabel{sub:theorem-4i}
|
||||
|
||||
\begin{theorem}[4I]
|
||||
For natural numbers $m$ and $n$,
|
||||
For \nameref{ref:natural-number}s $m$ and $n$,
|
||||
\begin{align}
|
||||
m + 0 & = m, \hyperlabel{sub:theorem-4i-eq1} \\
|
||||
m + n^+ & = (m + n)^+. \hyperlabel{sub:theorem-4i-eq2}
|
||||
|
|
@ -6328,7 +6326,7 @@
|
|||
\hyperlabel{sub:theorem-4k-1}
|
||||
|
||||
\begin{theorem}[4K-1]
|
||||
Associative law for addition.
|
||||
Associative law for \nameref{ref:addition}.
|
||||
For $m, n, p \in \omega$, $$m + (n + p) = (m + n) + p.$$
|
||||
\end{theorem}
|
||||
|
||||
|
|
@ -6385,7 +6383,7 @@
|
|||
\hyperlabel{sub:theorem-4k-2}
|
||||
|
||||
\begin{theorem}[4K-2]
|
||||
Commutative law for addition.
|
||||
Commutative law for \nameref{ref:addition}.
|
||||
For $m, n \in \omega$, $$m + n = n + m.$$
|
||||
\end{theorem}
|
||||
|
||||
|
|
@ -6716,7 +6714,7 @@
|
|||
\hyperlabel{sub:theorem-4k-5}
|
||||
|
||||
\begin{theorem}[4K-5]
|
||||
Commutative law for multiplication.
|
||||
Commutative law for \nameref{ref:multiplication}.
|
||||
For $m, n \in \omega$, $$m \cdot n = n \cdot m.$$
|
||||
\end{theorem}
|
||||
|
||||
|
|
@ -6777,7 +6775,7 @@
|
|||
\hyperlabel{sub:theorem-4k-4}
|
||||
|
||||
\begin{theorem}[4K-4]
|
||||
Associative law for multiplication.
|
||||
Associative law for \nameref{ref:multiplication}.
|
||||
For $m, n, p \in \omega$, $$m \cdot (n \cdot p) = (m \cdot n) \cdot p.$$
|
||||
\end{theorem}
|
||||
|
||||
|
|
@ -6859,7 +6857,8 @@
|
|||
\hyperlabel{sub:members-natural-numbers}
|
||||
|
||||
\begin{lemma}
|
||||
Every natural number is the set of all smaller natural numbers.
|
||||
Every \nameref{ref:natural-number} is the set of all smaller natural
|
||||
numbers.
|
||||
\end{lemma}
|
||||
|
||||
\begin{proof}
|
||||
|
|
@ -6874,7 +6873,8 @@
|
|||
\hyperlabel{sub:lemma-4l-a}
|
||||
|
||||
\begin{lemma}[4L(a)]
|
||||
For any natural numbers $m$ and $n$, $$m \in n \iff m^+ \in n^+.$$
|
||||
For any \nameref{ref:natural-number}s $m$ and $n$,
|
||||
$$m \in n \iff m^+ \in n^+.$$
|
||||
\end{lemma}
|
||||
|
||||
\lean{Std/Data/Nat/Lemmas}{Nat.succ\_lt\_succ\_iff}
|
||||
|
|
@ -6940,7 +6940,7 @@
|
|||
\hyperlabel{sub:lemma-4l-b}
|
||||
|
||||
\begin{lemma}[4L(b)]
|
||||
No natural number is a member of itself.
|
||||
No \nameref{ref:natural-number} is a member of itself.
|
||||
\end{lemma}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
|
|
@ -6987,7 +6987,7 @@
|
|||
\hyperlabel{sub:zero-least-natural-number}
|
||||
|
||||
\begin{lemma}
|
||||
For every natural number $n \neq 0$, $0 \in n$.
|
||||
For every \nameref{ref:natural-number} $n \neq 0$, $0 \in n$.
|
||||
\end{lemma}
|
||||
|
||||
\code{Bookshelf/Enderton/Set/Chapter\_4}
|
||||
|
|
@ -7032,8 +7032,8 @@
|
|||
\hyperlabel{sub:trichotomy-law-natural-numbers}
|
||||
|
||||
\begin{theorem}
|
||||
For any natural numbers $m$ and $n$, exactly one of the three conditions
|
||||
$$m \in n, \quad m = n, \quad n \in m$$ holds.
|
||||
For any \nameref{ref:natural-number}s $m$ and $n$, exactly one of the three
|
||||
conditions $$m \in n, \quad m = n, \quad n \in m$$ holds.
|
||||
\end{theorem}
|
||||
|
||||
\lean{Mathlib/Order/RelClasses}{IsAsymm}
|
||||
|
|
@ -7112,12 +7112,12 @@
|
|||
\hyperlabel{sub:linear-ordering-natural-numbers}
|
||||
|
||||
\begin{theorem}
|
||||
Relation
|
||||
\nameref{ref:relation}
|
||||
\begin{equation}
|
||||
\hyperlabel{sub:linear-ordering-natural-numbers-eq1}
|
||||
\in_\omega = \{\tuple{m, n} \in \omega \times \omega \mid m \in n\}
|
||||
\end{equation}
|
||||
is a linear ordering on $\omega$.
|
||||
is a \nameref{ref:linear-ordering} on $\omega$.
|
||||
\end{theorem}
|
||||
|
||||
\lean{Mathlib/Init/Algebra/Order}
|
||||
|
|
@ -7125,9 +7125,8 @@
|
|||
|
||||
\begin{proof}
|
||||
|
||||
By definition, \eqref{sub:linear-ordering-natural-numbers-eq1} is a
|
||||
\nameref{ref:linear-ordering} on $\omega$ if it is (i) transitive and
|
||||
(ii) trichotomous.
|
||||
By definition, \eqref{sub:linear-ordering-natural-numbers-eq1} is a linear
|
||||
ordering on $\omega$ if it is (i) transitive and (ii) trichotomous.
|
||||
|
||||
\paragraph{(i)}%
|
||||
|
||||
|
|
@ -8510,7 +8509,7 @@
|
|||
\hyperlabel{sub:theorem-6b}
|
||||
|
||||
\begin{theorem}[6B]
|
||||
No set is equinumerous to its powerset.
|
||||
No set is \nameref{ref:equinumerous} to its powerset.
|
||||
\end{theorem}
|
||||
|
||||
\begin{proof}
|
||||
|
|
|
|||
|
|
@ -15,6 +15,8 @@ Relations and Functions
|
|||
|
||||
namespace Enderton.Set.Chapter_3
|
||||
|
||||
open Set.Relation
|
||||
|
||||
/-- #### Lemma 3B
|
||||
|
||||
If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`.
|
||||
|
|
@ -25,203 +27,6 @@ lemma lemma_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
|
|||
have hxys : {x, y} ⊆ C := Set.mem_mem_imp_pair_subset hx hy
|
||||
exact Set.mem_mem_imp_pair_subset hxs hxys
|
||||
|
||||
/-- #### Exercise 3.1
|
||||
|
||||
Suppose that we attempted to generalize the Kuratowski definitions of ordered
|
||||
pairs to ordered triples by defining
|
||||
```
|
||||
⟨x, y, z⟩* = {{x}, {x, y}, {x, y, z}}.open Set
|
||||
|
||||
```
|
||||
Show that this definition is unsuccessful by giving examples of objects `u`,
|
||||
`v`, `w`, `x`, `y`, `z` with `⟨x, y, z⟩* = ⟨u, v, w⟩*` but with either `y ≠ v`
|
||||
or `z ≠ w` (or both).
|
||||
-/
|
||||
theorem exercise_3_1 {x y z u v w : ℕ}
|
||||
(hx : x = 1) (hy : y = 1) (hz : z = 2)
|
||||
(hu : u = 1) (hv : v = 2) (hw : w = 2)
|
||||
: ({{x}, {x, y}, {x, y, z}} : Set (Set ℕ)) = {{u}, {u, v}, {u, v, w}}
|
||||
∧ y ≠ v := by
|
||||
apply And.intro
|
||||
· rw [hx, hy, hz, hu, hv, hw]
|
||||
simp
|
||||
· rw [hy, hv]
|
||||
simp only
|
||||
|
||||
/-- #### Exercise 3.2a
|
||||
|
||||
Show that `A × (B ∪ C) = (A × B) ∪ (A × C)`.
|
||||
-/
|
||||
theorem exercise_3_2a {A : Set α} {B C : Set β}
|
||||
: Set.prod A (B ∪ C) = (Set.prod A B) ∪ (Set.prod A C) := by
|
||||
calc Set.prod A (B ∪ C)
|
||||
_ = { p | p.1 ∈ A ∧ p.2 ∈ B ∪ C } := rfl
|
||||
_ = { p | p.1 ∈ A ∧ (p.2 ∈ B ∨ p.2 ∈ C) } := rfl
|
||||
_ = { p | (p.1 ∈ A ∧ p.2 ∈ B) ∨ (p.1 ∈ A ∧ p.2 ∈ C) } := by
|
||||
ext x
|
||||
rw [Set.mem_setOf_eq]
|
||||
conv => lhs; rw [and_or_left]
|
||||
_ = { p | p ∈ Set.prod A B ∨ (p ∈ Set.prod A C) } := rfl
|
||||
_ = (Set.prod A B) ∪ (Set.prod A C) := rfl
|
||||
|
||||
/-- #### Exercise 3.2b
|
||||
|
||||
Show that if `A × B = A × C` and `A ≠ ∅`, then `B = C`.
|
||||
-/
|
||||
theorem exercise_3_2b {A : Set α} {B C : Set β}
|
||||
(h : Set.prod A B = Set.prod A C) (hA : Set.Nonempty A)
|
||||
: B = C := by
|
||||
by_cases hB : Set.Nonempty B
|
||||
· rw [Set.Subset.antisymm_iff]
|
||||
have ⟨a, ha⟩ := hA
|
||||
apply And.intro
|
||||
· show ∀ t, t ∈ B → t ∈ C
|
||||
intro t ht
|
||||
have : (a, t) ∈ Set.prod A B := ⟨ha, ht⟩
|
||||
rw [h] at this
|
||||
exact this.right
|
||||
· show ∀ t, t ∈ C → t ∈ B
|
||||
intro t ht
|
||||
have : (a, t) ∈ Set.prod A C := ⟨ha, ht⟩
|
||||
rw [← h] at this
|
||||
exact this.right
|
||||
· have nB : B = ∅ := Set.not_nonempty_iff_eq_empty.mp hB
|
||||
rw [nB, Set.prod_right_emptyset_eq_emptyset, Set.ext_iff] at h
|
||||
rw [nB]
|
||||
by_contra nC
|
||||
have ⟨a, ha⟩ := hA
|
||||
have ⟨c, hc⟩ := Set.nonempty_iff_ne_empty.mpr (Ne.symm nC)
|
||||
exact (h (a, c)).mpr ⟨ha, hc⟩
|
||||
|
||||
/-- #### Exercise 3.3
|
||||
|
||||
Show that `A × ⋃ 𝓑 = ⋃ {A × X | X ∈ 𝓑}`.
|
||||
-/
|
||||
theorem exercise_3_3 {A : Set (Set α)} {𝓑 : Set (Set β)}
|
||||
: Set.prod A (⋃₀ 𝓑) = ⋃₀ {Set.prod A X | X ∈ 𝓑} := by
|
||||
calc Set.prod A (⋃₀ 𝓑)
|
||||
_ = { p | p.1 ∈ A ∧ p.2 ∈ ⋃₀ 𝓑} := rfl
|
||||
_ = { p | p.1 ∈ A ∧ ∃ b ∈ 𝓑, p.2 ∈ b } := rfl
|
||||
_ = { p | ∃ b ∈ 𝓑, p.1 ∈ A ∧ p.2 ∈ b } := by
|
||||
ext x
|
||||
rw [Set.mem_setOf_eq]
|
||||
apply Iff.intro
|
||||
· intro ⟨h₁, b, h₂⟩
|
||||
exact ⟨b, h₂.left, h₁, h₂.right⟩
|
||||
· intro ⟨b, h₁, h₂, h₃⟩
|
||||
exact ⟨h₂, b, h₁, h₃⟩
|
||||
_ = ⋃₀ { Set.prod A p | p ∈ 𝓑 } := by
|
||||
ext x
|
||||
rw [Set.mem_setOf_eq]
|
||||
unfold Set.sUnion sSup Set.instSupSetSet
|
||||
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and]
|
||||
apply Iff.intro
|
||||
· intro ⟨b, h₁, h₂, h₃⟩
|
||||
exact ⟨b, h₁, h₂, h₃⟩
|
||||
· intro ⟨b, h₁, h₂, h₃⟩
|
||||
exact ⟨b, h₁, h₂, h₃⟩
|
||||
|
||||
/-- #### Exercise 3.5a
|
||||
|
||||
Assume that `A` and `B` are given sets, and show that there exists a set `C`
|
||||
such that for any `y`,
|
||||
```
|
||||
y ∈ C ↔ y = {x} × B for some x in A.
|
||||
```
|
||||
In other words, show that `{{x} × B | x ∈ A}` is a set.
|
||||
-/
|
||||
theorem exercise_3_5a {A : Set α} {B : Set β}
|
||||
: ∃ C : Set (Set (α × β)),
|
||||
y ∈ C ↔ ∃ x ∈ A, y = Set.prod {x} B := by
|
||||
let C := {y ∈ 𝒫 (Set.prod A B) | ∃ a ∈ A, ∀ x, (x ∈ y ↔ ∃ b ∈ B, x = (a, b))}
|
||||
refine ⟨C, ?_⟩
|
||||
apply Iff.intro
|
||||
· intro hC
|
||||
simp only [Set.mem_setOf_eq] at hC
|
||||
have ⟨_, ⟨a, ⟨ha, h⟩⟩⟩ := hC
|
||||
refine ⟨a, ⟨ha, ?_⟩⟩
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hxy
|
||||
unfold Set.prod
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
|
||||
have ⟨b, ⟨hb, hx⟩⟩ := (h x).mp hxy
|
||||
rw [Prod.ext_iff] at hx
|
||||
simp only at hx
|
||||
rw [← hx.right] at hb
|
||||
exact ⟨hx.left, hb⟩
|
||||
· intro hx
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hx
|
||||
have := (h (a, x.snd)).mpr ⟨x.snd, ⟨hx.right, rfl⟩⟩
|
||||
have hxab : x = (a, x.snd) := by
|
||||
ext <;> simp
|
||||
exact hx.left
|
||||
rwa [← hxab] at this
|
||||
· intro ⟨x, ⟨hx, hy⟩⟩
|
||||
show y ∈ 𝒫 Set.prod A B ∧ ∃ a, a ∈ A ∧
|
||||
∀ (x : α × β), x ∈ y ↔ ∃ b, b ∈ B ∧ x = (a, b)
|
||||
apply And.intro
|
||||
· simp only [Set.mem_powerset_iff]
|
||||
rw [hy]
|
||||
unfold Set.prod
|
||||
simp only [
|
||||
Set.mem_singleton_iff,
|
||||
Set.setOf_subset_setOf,
|
||||
and_imp,
|
||||
Prod.forall
|
||||
]
|
||||
intro a b ha hb
|
||||
exact ⟨by rw [ha]; exact hx, hb⟩
|
||||
· refine ⟨x, ⟨hx, ?_⟩⟩
|
||||
intro p
|
||||
apply Iff.intro
|
||||
· intro hab
|
||||
rw [hy] at hab
|
||||
unfold Set.prod at hab
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hab
|
||||
exact ⟨p.2, ⟨hab.right, by ext; exact hab.left; simp⟩⟩
|
||||
· intro ⟨b, ⟨hb, hab⟩⟩
|
||||
rw [hy]
|
||||
unfold Set.prod
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
|
||||
rw [Prod.ext_iff] at hab
|
||||
simp only at hab
|
||||
rw [hab.right]
|
||||
exact ⟨hab.left, hb⟩
|
||||
|
||||
/-- #### Exercise 3.5b
|
||||
|
||||
With `A`, `B`, and `C` as above, show that `A × B = ∪ C`.
|
||||
-/
|
||||
theorem exercise_3_5b {A : Set α} (B : Set β)
|
||||
: Set.prod A B = ⋃₀ {Set.prod ({x} : Set α) B | x ∈ A} := by
|
||||
rw [Set.Subset.antisymm_iff]
|
||||
apply And.intro
|
||||
· show ∀ t, t ∈ Set.prod A B → t ∈ ⋃₀ {Set.prod {x} B | x ∈ A}
|
||||
intro t h
|
||||
simp only [Set.mem_setOf_eq] at h
|
||||
unfold Set.sUnion sSup Set.instSupSetSet
|
||||
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and]
|
||||
unfold Set.prod at h
|
||||
simp only [Set.mem_setOf_eq] at h
|
||||
refine ⟨t.fst, ⟨h.left, ?_⟩⟩
|
||||
unfold Set.prod
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq, true_and]
|
||||
exact h.right
|
||||
· show ∀ t, t ∈ ⋃₀ {Set.prod {x} B | x ∈ A} → t ∈ Set.prod A B
|
||||
unfold Set.prod
|
||||
intro t ht
|
||||
simp only [
|
||||
Set.mem_singleton_iff,
|
||||
Set.mem_sUnion,
|
||||
Set.mem_setOf_eq,
|
||||
exists_exists_and_eq_and
|
||||
] at ht
|
||||
have ⟨a, ⟨h, ⟨ha, hb⟩⟩⟩ := ht
|
||||
simp only [Set.mem_setOf_eq]
|
||||
rw [← ha] at h
|
||||
exact ⟨h, hb⟩
|
||||
|
||||
/-- #### Theorem 3D
|
||||
|
||||
If `⟨x, y⟩ ∈ A`, then `x` and `y` belong to `⋃ ⋃ A`.
|
||||
|
|
@ -234,209 +39,6 @@ theorem theorem_3d {A : Set (Set (Set α))} (h : OrderedPair x y ∈ A)
|
|||
have : {x, y} ⊆ ⋃₀ ⋃₀ A := Chapter_2.exercise_2_3 {x, y} hq
|
||||
exact ⟨this (by simp), this (by simp)⟩
|
||||
|
||||
section Relation
|
||||
|
||||
open Set.Relation
|
||||
|
||||
/-- #### Exercise 3.6
|
||||
|
||||
Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`.
|
||||
-/
|
||||
theorem exercise_3_6 {A : Set.HRelation α β}
|
||||
: A ⊆ Set.prod (dom A) (ran A) := by
|
||||
show ∀ t, t ∈ A → t ∈ Set.prod (Prod.fst '' A) (Prod.snd '' A)
|
||||
intro (a, b) ht
|
||||
unfold Set.prod
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Prod.exists,
|
||||
exists_and_right,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq
|
||||
]
|
||||
exact ⟨⟨b, ht⟩, ⟨a, ht⟩⟩
|
||||
|
||||
/-- #### Exercise 3.7
|
||||
|
||||
Show that if `R` is a relation, then `fld R = ⋃ ⋃ R`.
|
||||
-/
|
||||
theorem exercise_3_7 {R : Set.Relation α}
|
||||
: R.fld = ⋃₀ ⋃₀ R.toOrderedPairs := by
|
||||
let img := R.toOrderedPairs
|
||||
rw [Set.Subset.antisymm_iff]
|
||||
apply And.intro
|
||||
· show ∀ x, x ∈ R.fld → x ∈ ⋃₀ ⋃₀ img
|
||||
intro x hx
|
||||
apply Or.elim hx
|
||||
· intro hd
|
||||
unfold Set.Relation.dom Prod.fst at hd
|
||||
simp only [
|
||||
Set.mem_image, Prod.exists, exists_and_right, exists_eq_right
|
||||
] at hd
|
||||
have ⟨y, hp⟩ := hd
|
||||
have hm : OrderedPair x y ∈ Set.image (fun p => OrderedPair p.1 p.2) R := by
|
||||
unfold Set.image
|
||||
simp only [Prod.exists, Set.mem_setOf_eq]
|
||||
exact ⟨x, ⟨y, ⟨hp, rfl⟩⟩⟩
|
||||
unfold OrderedPair at hm
|
||||
have : {x} ∈ ⋃₀ img := Chapter_2.exercise_2_3 {{x}, {x, y}} hm (by simp)
|
||||
exact (Chapter_2.exercise_2_3 {x} this) (show x ∈ {x} by rfl)
|
||||
· intro hr
|
||||
unfold Set.Relation.ran Prod.snd at hr
|
||||
simp only [Set.mem_image, Prod.exists, exists_eq_right] at hr
|
||||
have ⟨t, ht⟩ := hr
|
||||
have hm : OrderedPair t x ∈ Set.image (fun p => OrderedPair p.1 p.2) R := by
|
||||
simp only [Set.mem_image, Prod.exists]
|
||||
exact ⟨t, ⟨x, ⟨ht, rfl⟩⟩⟩
|
||||
unfold OrderedPair at hm
|
||||
have : {t, x} ∈ ⋃₀ img := Chapter_2.exercise_2_3 {{t}, {t, x}} hm
|
||||
(show {t, x} ∈ {{t}, {t, x}} by simp)
|
||||
exact Chapter_2.exercise_2_3 {t, x} this (show x ∈ {t, x} by simp)
|
||||
|
||||
· show ∀ t, t ∈ ⋃₀ ⋃₀ img → t ∈ Set.Relation.fld R
|
||||
intro t ht
|
||||
have ⟨T, hT⟩ : ∃ T ∈ ⋃₀ img, t ∈ T := ht
|
||||
have ⟨T', hT'⟩ : ∃ T' ∈ img, T ∈ T' := hT.left
|
||||
dsimp only at hT'
|
||||
unfold Set.Relation.toOrderedPairs at hT'
|
||||
simp only [Set.mem_image, Prod.exists] at hT'
|
||||
have ⟨x, ⟨y, ⟨p, hp⟩⟩⟩ := hT'.left
|
||||
have hr := hT'.right
|
||||
rw [← hp] at hr
|
||||
unfold OrderedPair at hr
|
||||
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at hr
|
||||
|
||||
-- Use `exercise_6_6` to prove that if `t = x` then `t ∈ dom R` and if
|
||||
-- `t = y` then `t ∈ ran R`.
|
||||
have hxy_mem : t = x ∨ t = y → t ∈ Set.Relation.fld R := by
|
||||
intro ht
|
||||
have hz : R ⊆ Set.prod (dom R) (ran R) := exercise_3_6
|
||||
have : (x, y) ∈ Set.prod (dom R) (ran R) := hz p
|
||||
unfold Set.prod at this
|
||||
simp at this
|
||||
apply Or.elim ht
|
||||
· intro ht'
|
||||
rw [← ht'] at this
|
||||
exact Or.inl this.left
|
||||
· intro ht'
|
||||
rw [← ht'] at this
|
||||
exact Or.inr this.right
|
||||
|
||||
-- Eliminate `T = {x} ∨ T = {x, y}`.
|
||||
apply Or.elim hr
|
||||
· intro hx
|
||||
have := hT.right
|
||||
rw [hx] at this
|
||||
simp only [Set.mem_singleton_iff] at this
|
||||
exact hxy_mem (Or.inl this)
|
||||
· intro hxy
|
||||
have := hT.right
|
||||
rw [hxy] at this
|
||||
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this
|
||||
exact hxy_mem this
|
||||
|
||||
/-- #### Exercise 3.8 (i)
|
||||
|
||||
Show that for any set `𝓐`:
|
||||
```
|
||||
dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_3_8_i {A : Set (Set.HRelation α β)}
|
||||
: dom (⋃₀ A) = ⋃₀ { dom R | R ∈ A } := by
|
||||
ext x
|
||||
unfold dom Prod.fst
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Set.mem_sUnion,
|
||||
Prod.exists,
|
||||
exists_and_right,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq,
|
||||
exists_exists_and_eq_and
|
||||
]
|
||||
apply Iff.intro
|
||||
· intro ⟨y, t, ht, hx⟩
|
||||
exact ⟨t, ht, y, hx⟩
|
||||
· intro ⟨t, ht, y, hx⟩
|
||||
exact ⟨y, t, ht, hx⟩
|
||||
|
||||
/-- #### Exercise 3.8 (ii)
|
||||
|
||||
Show that for any set `𝓐`:
|
||||
```
|
||||
ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_3_8_ii {A : Set (Set.HRelation α β)}
|
||||
: ran (⋃₀ A) = ⋃₀ { ran R | R ∈ A } := by
|
||||
ext x
|
||||
unfold ran Prod.snd
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Set.mem_sUnion,
|
||||
Prod.exists,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq,
|
||||
exists_exists_and_eq_and
|
||||
]
|
||||
apply Iff.intro
|
||||
· intro ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
|
||||
exact ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
|
||||
· intro ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
|
||||
exact ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
|
||||
|
||||
/-- #### Exercise 3.9 (i)
|
||||
|
||||
Discuss the result of replacing the union operation by the intersection
|
||||
operation in the preceding problem.
|
||||
```
|
||||
dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_3_9_i {A : Set (Set.HRelation α β)}
|
||||
: dom (⋂₀ A) ⊆ ⋂₀ { dom R | R ∈ A } := by
|
||||
show ∀ x, x ∈ dom (⋂₀ A) → x ∈ ⋂₀ { dom R | R ∈ A }
|
||||
unfold dom Prod.fst
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Set.mem_sInter,
|
||||
Prod.exists,
|
||||
exists_and_right,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq,
|
||||
forall_exists_index,
|
||||
and_imp,
|
||||
forall_apply_eq_imp_iff₂
|
||||
]
|
||||
intro _ y hy R hR
|
||||
exact ⟨y, hy R hR⟩
|
||||
|
||||
/-- #### Exercise 3.9 (ii)
|
||||
|
||||
Discuss the result of replacing the union operation by the intersection
|
||||
operation in the preceding problem.
|
||||
```
|
||||
ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_3_9_ii {A : Set (Set.HRelation α β)}
|
||||
: ran (⋂₀ A) ⊆ ⋂₀ { ran R | R ∈ A } := by
|
||||
show ∀ x, x ∈ ran (⋂₀ A) → x ∈ ⋂₀ { ran R | R ∈ A }
|
||||
unfold ran Prod.snd
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Set.mem_sInter,
|
||||
Prod.exists,
|
||||
exists_and_right,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq,
|
||||
forall_exists_index,
|
||||
and_imp,
|
||||
forall_apply_eq_imp_iff₂
|
||||
]
|
||||
intro _ y hy R hR
|
||||
exact ⟨y, hy R hR⟩
|
||||
|
||||
/-- #### Theorem 3G (i)
|
||||
|
||||
|
|
@ -729,6 +331,428 @@ theorem corollary_3l_iii {G : Set.HRelation β α} {A B : Set α}
|
|||
single_valued_self_iff_single_rooted_inv.mp hG
|
||||
exact (theorem_3k_c_ii hG').symm
|
||||
|
||||
/-- #### Theorem 3R
|
||||
|
||||
Let `R` be a linear ordering on `A`.
|
||||
|
||||
(i) There is no `x` for which `xRx`.
|
||||
(ii) For distinct `x` and `y` in `A`, either `xRy` or `yRx`.
|
||||
-/
|
||||
theorem theorem_3r {R : Rel α α} (hR : IsStrictTotalOrder α R)
|
||||
: (∀ x : α, ¬ R x x) ∧ (∀ x y : α, x ≠ y → R x y ∨ R y x) := by
|
||||
apply And.intro
|
||||
· exact hR.irrefl
|
||||
· intro x y h
|
||||
apply Or.elim (hR.trichotomous x y)
|
||||
· intro h₁
|
||||
left
|
||||
exact h₁
|
||||
· intro h₁
|
||||
apply Or.elim h₁
|
||||
· intro h₂
|
||||
exact absurd h₂ h
|
||||
· intro h₂
|
||||
right
|
||||
exact h₂
|
||||
|
||||
/-- #### Exercise 3.1
|
||||
|
||||
Suppose that we attempted to generalize the Kuratowski definitions of ordered
|
||||
pairs to ordered triples by defining
|
||||
```
|
||||
⟨x, y, z⟩* = {{x}, {x, y}, {x, y, z}}.open Set
|
||||
|
||||
```
|
||||
Show that this definition is unsuccessful by giving examples of objects `u`,
|
||||
`v`, `w`, `x`, `y`, `z` with `⟨x, y, z⟩* = ⟨u, v, w⟩*` but with either `y ≠ v`
|
||||
or `z ≠ w` (or both).
|
||||
-/
|
||||
theorem exercise_3_1 {x y z u v w : ℕ}
|
||||
(hx : x = 1) (hy : y = 1) (hz : z = 2)
|
||||
(hu : u = 1) (hv : v = 2) (hw : w = 2)
|
||||
: ({{x}, {x, y}, {x, y, z}} : Set (Set ℕ)) = {{u}, {u, v}, {u, v, w}}
|
||||
∧ y ≠ v := by
|
||||
apply And.intro
|
||||
· rw [hx, hy, hz, hu, hv, hw]
|
||||
simp
|
||||
· rw [hy, hv]
|
||||
simp only
|
||||
|
||||
/-- #### Exercise 3.2a
|
||||
|
||||
Show that `A × (B ∪ C) = (A × B) ∪ (A × C)`.
|
||||
-/
|
||||
theorem exercise_3_2a {A : Set α} {B C : Set β}
|
||||
: Set.prod A (B ∪ C) = (Set.prod A B) ∪ (Set.prod A C) := by
|
||||
calc Set.prod A (B ∪ C)
|
||||
_ = { p | p.1 ∈ A ∧ p.2 ∈ B ∪ C } := rfl
|
||||
_ = { p | p.1 ∈ A ∧ (p.2 ∈ B ∨ p.2 ∈ C) } := rfl
|
||||
_ = { p | (p.1 ∈ A ∧ p.2 ∈ B) ∨ (p.1 ∈ A ∧ p.2 ∈ C) } := by
|
||||
ext x
|
||||
rw [Set.mem_setOf_eq]
|
||||
conv => lhs; rw [and_or_left]
|
||||
_ = { p | p ∈ Set.prod A B ∨ (p ∈ Set.prod A C) } := rfl
|
||||
_ = (Set.prod A B) ∪ (Set.prod A C) := rfl
|
||||
|
||||
/-- #### Exercise 3.2b
|
||||
|
||||
Show that if `A × B = A × C` and `A ≠ ∅`, then `B = C`.
|
||||
-/
|
||||
theorem exercise_3_2b {A : Set α} {B C : Set β}
|
||||
(h : Set.prod A B = Set.prod A C) (hA : Set.Nonempty A)
|
||||
: B = C := by
|
||||
by_cases hB : Set.Nonempty B
|
||||
· rw [Set.Subset.antisymm_iff]
|
||||
have ⟨a, ha⟩ := hA
|
||||
apply And.intro
|
||||
· show ∀ t, t ∈ B → t ∈ C
|
||||
intro t ht
|
||||
have : (a, t) ∈ Set.prod A B := ⟨ha, ht⟩
|
||||
rw [h] at this
|
||||
exact this.right
|
||||
· show ∀ t, t ∈ C → t ∈ B
|
||||
intro t ht
|
||||
have : (a, t) ∈ Set.prod A C := ⟨ha, ht⟩
|
||||
rw [← h] at this
|
||||
exact this.right
|
||||
· have nB : B = ∅ := Set.not_nonempty_iff_eq_empty.mp hB
|
||||
rw [nB, Set.prod_right_emptyset_eq_emptyset, Set.ext_iff] at h
|
||||
rw [nB]
|
||||
by_contra nC
|
||||
have ⟨a, ha⟩ := hA
|
||||
have ⟨c, hc⟩ := Set.nonempty_iff_ne_empty.mpr (Ne.symm nC)
|
||||
exact (h (a, c)).mpr ⟨ha, hc⟩
|
||||
|
||||
/-- #### Exercise 3.3
|
||||
|
||||
Show that `A × ⋃ 𝓑 = ⋃ {A × X | X ∈ 𝓑}`.
|
||||
-/
|
||||
theorem exercise_3_3 {A : Set (Set α)} {𝓑 : Set (Set β)}
|
||||
: Set.prod A (⋃₀ 𝓑) = ⋃₀ {Set.prod A X | X ∈ 𝓑} := by
|
||||
calc Set.prod A (⋃₀ 𝓑)
|
||||
_ = { p | p.1 ∈ A ∧ p.2 ∈ ⋃₀ 𝓑} := rfl
|
||||
_ = { p | p.1 ∈ A ∧ ∃ b ∈ 𝓑, p.2 ∈ b } := rfl
|
||||
_ = { p | ∃ b ∈ 𝓑, p.1 ∈ A ∧ p.2 ∈ b } := by
|
||||
ext x
|
||||
rw [Set.mem_setOf_eq]
|
||||
apply Iff.intro
|
||||
· intro ⟨h₁, b, h₂⟩
|
||||
exact ⟨b, h₂.left, h₁, h₂.right⟩
|
||||
· intro ⟨b, h₁, h₂, h₃⟩
|
||||
exact ⟨h₂, b, h₁, h₃⟩
|
||||
_ = ⋃₀ { Set.prod A p | p ∈ 𝓑 } := by
|
||||
ext x
|
||||
rw [Set.mem_setOf_eq]
|
||||
unfold Set.sUnion sSup Set.instSupSetSet
|
||||
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and]
|
||||
apply Iff.intro
|
||||
· intro ⟨b, h₁, h₂, h₃⟩
|
||||
exact ⟨b, h₁, h₂, h₃⟩
|
||||
· intro ⟨b, h₁, h₂, h₃⟩
|
||||
exact ⟨b, h₁, h₂, h₃⟩
|
||||
|
||||
/-- #### Exercise 3.5a
|
||||
|
||||
Assume that `A` and `B` are given sets, and show that there exists a set `C`
|
||||
such that for any `y`,
|
||||
```
|
||||
y ∈ C ↔ y = {x} × B for some x in A.
|
||||
```
|
||||
In other words, show that `{{x} × B | x ∈ A}` is a set.
|
||||
-/
|
||||
theorem exercise_3_5a {A : Set α} {B : Set β}
|
||||
: ∃ C : Set (Set (α × β)),
|
||||
y ∈ C ↔ ∃ x ∈ A, y = Set.prod {x} B := by
|
||||
let C := {y ∈ 𝒫 (Set.prod A B) | ∃ a ∈ A, ∀ x, (x ∈ y ↔ ∃ b ∈ B, x = (a, b))}
|
||||
refine ⟨C, ?_⟩
|
||||
apply Iff.intro
|
||||
· intro hC
|
||||
simp only [Set.mem_setOf_eq] at hC
|
||||
have ⟨_, ⟨a, ⟨ha, h⟩⟩⟩ := hC
|
||||
refine ⟨a, ⟨ha, ?_⟩⟩
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hxy
|
||||
unfold Set.prod
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
|
||||
have ⟨b, ⟨hb, hx⟩⟩ := (h x).mp hxy
|
||||
rw [Prod.ext_iff] at hx
|
||||
simp only at hx
|
||||
rw [← hx.right] at hb
|
||||
exact ⟨hx.left, hb⟩
|
||||
· intro hx
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hx
|
||||
have := (h (a, x.snd)).mpr ⟨x.snd, ⟨hx.right, rfl⟩⟩
|
||||
have hxab : x = (a, x.snd) := by
|
||||
ext <;> simp
|
||||
exact hx.left
|
||||
rwa [← hxab] at this
|
||||
· intro ⟨x, ⟨hx, hy⟩⟩
|
||||
show y ∈ 𝒫 Set.prod A B ∧ ∃ a, a ∈ A ∧
|
||||
∀ (x : α × β), x ∈ y ↔ ∃ b, b ∈ B ∧ x = (a, b)
|
||||
apply And.intro
|
||||
· simp only [Set.mem_powerset_iff]
|
||||
rw [hy]
|
||||
unfold Set.prod
|
||||
simp only [
|
||||
Set.mem_singleton_iff,
|
||||
Set.setOf_subset_setOf,
|
||||
and_imp,
|
||||
Prod.forall
|
||||
]
|
||||
intro a b ha hb
|
||||
exact ⟨by rw [ha]; exact hx, hb⟩
|
||||
· refine ⟨x, ⟨hx, ?_⟩⟩
|
||||
intro p
|
||||
apply Iff.intro
|
||||
· intro hab
|
||||
rw [hy] at hab
|
||||
unfold Set.prod at hab
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hab
|
||||
exact ⟨p.2, ⟨hab.right, by ext; exact hab.left; simp⟩⟩
|
||||
· intro ⟨b, ⟨hb, hab⟩⟩
|
||||
rw [hy]
|
||||
unfold Set.prod
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
|
||||
rw [Prod.ext_iff] at hab
|
||||
simp only at hab
|
||||
rw [hab.right]
|
||||
exact ⟨hab.left, hb⟩
|
||||
|
||||
/-- #### Exercise 3.5b
|
||||
|
||||
With `A`, `B`, and `C` as above, show that `A × B = ∪ C`.
|
||||
-/
|
||||
theorem exercise_3_5b {A : Set α} (B : Set β)
|
||||
: Set.prod A B = ⋃₀ {Set.prod ({x} : Set α) B | x ∈ A} := by
|
||||
rw [Set.Subset.antisymm_iff]
|
||||
apply And.intro
|
||||
· show ∀ t, t ∈ Set.prod A B → t ∈ ⋃₀ {Set.prod {x} B | x ∈ A}
|
||||
intro t h
|
||||
simp only [Set.mem_setOf_eq] at h
|
||||
unfold Set.sUnion sSup Set.instSupSetSet
|
||||
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and]
|
||||
unfold Set.prod at h
|
||||
simp only [Set.mem_setOf_eq] at h
|
||||
refine ⟨t.fst, ⟨h.left, ?_⟩⟩
|
||||
unfold Set.prod
|
||||
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq, true_and]
|
||||
exact h.right
|
||||
· show ∀ t, t ∈ ⋃₀ {Set.prod {x} B | x ∈ A} → t ∈ Set.prod A B
|
||||
unfold Set.prod
|
||||
intro t ht
|
||||
simp only [
|
||||
Set.mem_singleton_iff,
|
||||
Set.mem_sUnion,
|
||||
Set.mem_setOf_eq,
|
||||
exists_exists_and_eq_and
|
||||
] at ht
|
||||
have ⟨a, ⟨h, ⟨ha, hb⟩⟩⟩ := ht
|
||||
simp only [Set.mem_setOf_eq]
|
||||
rw [← ha] at h
|
||||
exact ⟨h, hb⟩
|
||||
|
||||
|
||||
/-- #### Exercise 3.6
|
||||
|
||||
Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`.
|
||||
-/
|
||||
theorem exercise_3_6 {A : Set.HRelation α β}
|
||||
: A ⊆ Set.prod (dom A) (ran A) := by
|
||||
show ∀ t, t ∈ A → t ∈ Set.prod (Prod.fst '' A) (Prod.snd '' A)
|
||||
intro (a, b) ht
|
||||
unfold Set.prod
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Prod.exists,
|
||||
exists_and_right,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq
|
||||
]
|
||||
exact ⟨⟨b, ht⟩, ⟨a, ht⟩⟩
|
||||
|
||||
/-- #### Exercise 3.7
|
||||
|
||||
Show that if `R` is a relation, then `fld R = ⋃ ⋃ R`.
|
||||
-/
|
||||
theorem exercise_3_7 {R : Set.Relation α}
|
||||
: R.fld = ⋃₀ ⋃₀ R.toOrderedPairs := by
|
||||
let img := R.toOrderedPairs
|
||||
rw [Set.Subset.antisymm_iff]
|
||||
apply And.intro
|
||||
· show ∀ x, x ∈ R.fld → x ∈ ⋃₀ ⋃₀ img
|
||||
intro x hx
|
||||
apply Or.elim hx
|
||||
· intro hd
|
||||
unfold Set.Relation.dom Prod.fst at hd
|
||||
simp only [
|
||||
Set.mem_image, Prod.exists, exists_and_right, exists_eq_right
|
||||
] at hd
|
||||
have ⟨y, hp⟩ := hd
|
||||
have hm : OrderedPair x y ∈ Set.image (fun p => OrderedPair p.1 p.2) R := by
|
||||
unfold Set.image
|
||||
simp only [Prod.exists, Set.mem_setOf_eq]
|
||||
exact ⟨x, ⟨y, ⟨hp, rfl⟩⟩⟩
|
||||
unfold OrderedPair at hm
|
||||
have : {x} ∈ ⋃₀ img := Chapter_2.exercise_2_3 {{x}, {x, y}} hm (by simp)
|
||||
exact (Chapter_2.exercise_2_3 {x} this) (show x ∈ {x} by rfl)
|
||||
· intro hr
|
||||
unfold Set.Relation.ran Prod.snd at hr
|
||||
simp only [Set.mem_image, Prod.exists, exists_eq_right] at hr
|
||||
have ⟨t, ht⟩ := hr
|
||||
have hm : OrderedPair t x ∈ Set.image (fun p => OrderedPair p.1 p.2) R := by
|
||||
simp only [Set.mem_image, Prod.exists]
|
||||
exact ⟨t, ⟨x, ⟨ht, rfl⟩⟩⟩
|
||||
unfold OrderedPair at hm
|
||||
have : {t, x} ∈ ⋃₀ img := Chapter_2.exercise_2_3 {{t}, {t, x}} hm
|
||||
(show {t, x} ∈ {{t}, {t, x}} by simp)
|
||||
exact Chapter_2.exercise_2_3 {t, x} this (show x ∈ {t, x} by simp)
|
||||
|
||||
· show ∀ t, t ∈ ⋃₀ ⋃₀ img → t ∈ Set.Relation.fld R
|
||||
intro t ht
|
||||
have ⟨T, hT⟩ : ∃ T ∈ ⋃₀ img, t ∈ T := ht
|
||||
have ⟨T', hT'⟩ : ∃ T' ∈ img, T ∈ T' := hT.left
|
||||
dsimp only at hT'
|
||||
unfold Set.Relation.toOrderedPairs at hT'
|
||||
simp only [Set.mem_image, Prod.exists] at hT'
|
||||
have ⟨x, ⟨y, ⟨p, hp⟩⟩⟩ := hT'.left
|
||||
have hr := hT'.right
|
||||
rw [← hp] at hr
|
||||
unfold OrderedPair at hr
|
||||
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at hr
|
||||
|
||||
-- Use `exercise_6_6` to prove that if `t = x` then `t ∈ dom R` and if
|
||||
-- `t = y` then `t ∈ ran R`.
|
||||
have hxy_mem : t = x ∨ t = y → t ∈ Set.Relation.fld R := by
|
||||
intro ht
|
||||
have hz : R ⊆ Set.prod (dom R) (ran R) := exercise_3_6
|
||||
have : (x, y) ∈ Set.prod (dom R) (ran R) := hz p
|
||||
unfold Set.prod at this
|
||||
simp at this
|
||||
apply Or.elim ht
|
||||
· intro ht'
|
||||
rw [← ht'] at this
|
||||
exact Or.inl this.left
|
||||
· intro ht'
|
||||
rw [← ht'] at this
|
||||
exact Or.inr this.right
|
||||
|
||||
-- Eliminate `T = {x} ∨ T = {x, y}`.
|
||||
apply Or.elim hr
|
||||
· intro hx
|
||||
have := hT.right
|
||||
rw [hx] at this
|
||||
simp only [Set.mem_singleton_iff] at this
|
||||
exact hxy_mem (Or.inl this)
|
||||
· intro hxy
|
||||
have := hT.right
|
||||
rw [hxy] at this
|
||||
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this
|
||||
exact hxy_mem this
|
||||
|
||||
/-- #### Exercise 3.8 (i)
|
||||
|
||||
Show that for any set `𝓐`:
|
||||
```
|
||||
dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_3_8_i {A : Set (Set.HRelation α β)}
|
||||
: dom (⋃₀ A) = ⋃₀ { dom R | R ∈ A } := by
|
||||
ext x
|
||||
unfold dom Prod.fst
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Set.mem_sUnion,
|
||||
Prod.exists,
|
||||
exists_and_right,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq,
|
||||
exists_exists_and_eq_and
|
||||
]
|
||||
apply Iff.intro
|
||||
· intro ⟨y, t, ht, hx⟩
|
||||
exact ⟨t, ht, y, hx⟩
|
||||
· intro ⟨t, ht, y, hx⟩
|
||||
exact ⟨y, t, ht, hx⟩
|
||||
|
||||
/-- #### Exercise 3.8 (ii)
|
||||
|
||||
Show that for any set `𝓐`:
|
||||
```
|
||||
ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_3_8_ii {A : Set (Set.HRelation α β)}
|
||||
: ran (⋃₀ A) = ⋃₀ { ran R | R ∈ A } := by
|
||||
ext x
|
||||
unfold ran Prod.snd
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Set.mem_sUnion,
|
||||
Prod.exists,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq,
|
||||
exists_exists_and_eq_and
|
||||
]
|
||||
apply Iff.intro
|
||||
· intro ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
|
||||
exact ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
|
||||
· intro ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
|
||||
exact ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
|
||||
|
||||
/-- #### Exercise 3.9 (i)
|
||||
|
||||
Discuss the result of replacing the union operation by the intersection
|
||||
operation in the preceding problem.
|
||||
```
|
||||
dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_3_9_i {A : Set (Set.HRelation α β)}
|
||||
: dom (⋂₀ A) ⊆ ⋂₀ { dom R | R ∈ A } := by
|
||||
show ∀ x, x ∈ dom (⋂₀ A) → x ∈ ⋂₀ { dom R | R ∈ A }
|
||||
unfold dom Prod.fst
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Set.mem_sInter,
|
||||
Prod.exists,
|
||||
exists_and_right,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq,
|
||||
forall_exists_index,
|
||||
and_imp,
|
||||
forall_apply_eq_imp_iff₂
|
||||
]
|
||||
intro _ y hy R hR
|
||||
exact ⟨y, hy R hR⟩
|
||||
|
||||
/-- #### Exercise 3.9 (ii)
|
||||
|
||||
Discuss the result of replacing the union operation by the intersection
|
||||
operation in the preceding problem.
|
||||
```
|
||||
ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_3_9_ii {A : Set (Set.HRelation α β)}
|
||||
: ran (⋂₀ A) ⊆ ⋂₀ { ran R | R ∈ A } := by
|
||||
show ∀ x, x ∈ ran (⋂₀ A) → x ∈ ⋂₀ { ran R | R ∈ A }
|
||||
unfold ran Prod.snd
|
||||
simp only [
|
||||
Set.mem_image,
|
||||
Set.mem_sInter,
|
||||
Prod.exists,
|
||||
exists_and_right,
|
||||
exists_eq_right,
|
||||
Set.mem_setOf_eq,
|
||||
forall_exists_index,
|
||||
and_imp,
|
||||
forall_apply_eq_imp_iff₂
|
||||
]
|
||||
intro _ y hy R hR
|
||||
exact ⟨y, hy R hR⟩
|
||||
|
||||
/-- #### Exercise 3.12
|
||||
|
||||
Assume that `f` and `g` are functions and show that
|
||||
|
|
@ -2166,32 +2190,6 @@ theorem exercise_3_41_a {Q : Set.Relation (ℝ × ℝ)}
|
|||
conv => right; rw [add_comm]
|
||||
exact this
|
||||
|
||||
end Relation
|
||||
|
||||
/-- #### Theorem 3R
|
||||
|
||||
Let `R` be a linear ordering on `A`.
|
||||
|
||||
(i) There is no `x` for which `xRx`.
|
||||
(ii) For distinct `x` and `y` in `A`, either `xRy` or `yRx`.
|
||||
-/
|
||||
theorem theorem_3r {R : Rel α α} (hR : IsStrictTotalOrder α R)
|
||||
: (∀ x : α, ¬ R x x) ∧ (∀ x y : α, x ≠ y → R x y ∨ R y x) := by
|
||||
apply And.intro
|
||||
· exact hR.irrefl
|
||||
· intro x y h
|
||||
apply Or.elim (hR.trichotomous x y)
|
||||
· intro h₁
|
||||
left
|
||||
exact h₁
|
||||
· intro h₁
|
||||
apply Or.elim h₁
|
||||
· intro h₂
|
||||
exact absurd h₂ h
|
||||
· intro h₂
|
||||
right
|
||||
exact h₂
|
||||
|
||||
/-- #### Exercise 3.43
|
||||
|
||||
Assume that `R` is a linear ordering on a set `A`. Show that `R⁻¹` is also a
|
||||
|
|
|
|||
|
|
@ -36,4 +36,50 @@ theorem theorem_6a_c (A : Set α) (B : Set β) (C : Set γ)
|
|||
: ∃ H, Set.BijOn H A C := by
|
||||
exact ⟨G ∘ F, Set.BijOn.comp hG hF⟩
|
||||
|
||||
/-- #### Exercise 6.1
|
||||
|
||||
Show that the equation
|
||||
```
|
||||
f(m, n) = 2ᵐ(2n + 1) - 1
|
||||
```
|
||||
defines a one-to-one correspondence between `ω × ω` and `ω`.
|
||||
-/
|
||||
theorem exercise_6_1
|
||||
: Function.Bijective (fun p : ℕ × ℕ => 2 ^ p.1 * (2 * p.2 + 1) - 1) := by
|
||||
sorry
|
||||
|
||||
/-- #### Exercise 6.2
|
||||
|
||||
Show that in Fig. 32 we have:
|
||||
```
|
||||
J(m, n) = [1 + 2 + ⋯ + (m + n)] + m
|
||||
= (1 / 2)[(m + n)² + 3m + n].
|
||||
```
|
||||
-/
|
||||
theorem exercise_6_2
|
||||
: Function.Bijective
|
||||
(fun p : ℕ × ℕ => (1 / 2) * ((p.1 + p.2) ^ 2 + 3 * p.1 + p.2)) := by
|
||||
sorry
|
||||
|
||||
/-- #### Exercise 6.3
|
||||
|
||||
Find a one-to-one correspondence between the open unit interval `(0, 1)` and `ℝ`
|
||||
that takes rationals to rationals and irrationals to irrationals.
|
||||
-/
|
||||
theorem exercise_6_3
|
||||
: True := by
|
||||
sorry
|
||||
|
||||
/-- #### Exercise 6.4
|
||||
|
||||
Construct a one-to-one correspondence between the closed unit interval
|
||||
```
|
||||
[0, 1] = {x ∈ ℝ | 0 ≤ x ≤ 1}
|
||||
```
|
||||
and the open unit interval `(0, 1)`.
|
||||
-/
|
||||
theorem exercise_6_4
|
||||
: ∃ F, Set.BijOn F (Set.Ioo 0 1) (@Set.univ ℝ) := by
|
||||
sorry
|
||||
|
||||
end Enderton.Set.Chapter_6
|
||||
|
|
|
|||
Loading…
Reference in New Issue