Enderton. Finish exercises 6.
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@ -2792,36 +2792,127 @@ Show that if $R$ is a relation, then $\fld{R} = \bigcup\bigcup R$.
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\end{proof}
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\subsection{\unverified{Exercise 6.8}}%
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\subsection{\verified{Exercise 6.8}}%
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\label{sub:exercise-6.8}
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Show that for any set $\mathscr{A}$:
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\begin{align}
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\dom{\bigcup{\mathscr{A}}}
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& = \bigcup\; \{ \dom{R} \mid R \in \mathscr{A} \},
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& = \bigcup\;\{ \dom{R} \mid R \in \mathscr{A} \},
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& \label{sub:exercise-6.8-eq1} \\
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\ran{\bigcup{\mathscr{A}}}
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& = \bigcup\; \{ \ran{R} \mid R \in \mathscr{A} \}.
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& = \bigcup\;\{ \ran{R} \mid R \in \mathscr{A} \}.
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& \label{sub:exercise-6.8-eq2}
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\end{align}
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_6\_8\_i}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_6\_8\_ii}
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We prove (i) \eqref{sub:exercise-6.8-eq1} and then (ii)
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\eqref{sub:exercise-6.8-eq2}.
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\paragraph{(i)}%
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Let $x \in \dom{\bigcup{\mathscr{A}}}$.
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By definition of a domain, there exists some $y$ such that
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$\left< x, y \right> \in \bigcup{\mathscr{A}}$.
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By definition of the union of sets,
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$\exists y, \exists R \in \mathscr{A}, \left< x, y \right> \in R$.
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Equivalently,
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$\exists R \in \mathscr{A}, \exists y, \left< x, y \right> \in R$.
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By another application of the definition of a domain,
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$\exists R \in \mathscr{A}, x \in \dom{R}$.
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By another application of the definition of the union of sets,
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$x \in \bigcup\;\{ \dom{R} \mid R \in \mathscr{A} \}$.
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Since membership of these two sets holds biconditionally, it follows
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\eqref{sub:exercise-6.8-eq1} holds.
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\paragraph{(ii)}%
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Let $x \in \ran{\bigcup{\mathscr{A}}}$.
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By definition of a range, there exists some $t$ such that
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$\left< t, x \right> \in \bigcup{\mathscr{A}}$.
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By definition of the union of sets,
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$\exists t, \exists R \in \mathscr{A}, \left< t, x \right> \in R$.
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Equivalently,
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$\exists R \in \mathscr{A}, \exists t, \left< t, x \right> \in R$.
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By another application of the definition of a range,
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$\exists R \in \mathscr{A}, x \in \ran{R}$.
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By another application of the definition of the union of sets,
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$x \in \bigcup\;\{ \ran{R} \mid R \in \mathscr{A} \}$.
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Since membership of these two sets holds biconditionally, it follows
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\eqref{sub:exercise-6.8-eq2} holds.
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\end{proof}
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\subsection{\unverified{Exercise 6.9}}%
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\subsection{\verified{Exercise 6.9}}%
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\label{sub:exercise-6.9}
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Discuss the result of replacing the union operation by the intersection
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operation in the preceding problem.
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\begin{proof}
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\begin{answer}
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TODO
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\statementpadding
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\end{proof}
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_6\_9\_i}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_6\_9\_ii}
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Replacing the union operation with the intersection problem produces the
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following relationships
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\begin{align}
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\dom{\bigcap{\mathscr{A}}}
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& \subseteq \bigcap\;\{ \dom{R} \mid R \in \mathscr{A} \},
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& \label{sub:exercise-6.9-eq1} \\
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\ran{\bigcap{\mathscr{A}}}
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& \subseteq \bigcap\;\{ \ran{R} \mid R \in \mathscr{A} \}.
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& \label{sub:exercise-6.9-eq2}
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\end{align}
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We prove (i) \eqref{sub:exercise-6.9-eq1} and then (ii)
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\eqref{sub:exercise-6.9-eq2}.
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\paragraph{(i)}%
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Let $x \in \dom{\bigcap{\mathscr{A}}}$.
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By definition of the domain of a set,
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$\exists y, \left< x, y \right> \in \bigcap{\mathscr{A}}$.
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By definition of the intersection of sets,
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$\exists y, \forall R \in \mathscr{A}, \left< x, y \right> \in R$.
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But this implies that
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$\forall R \in \mathscr{A}, \exists y, \left< x, y \right> \in R$.
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By another application of the definition of the domain of a set,
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$\forall R \in \mathscr{A}, x \in \dom{R}$.
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By another application of the intersection of sets,
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$x \in \bigcap\;\{ \dom{R} \mid R \in \mathscr{A} \}$.
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Thus \eqref{sub:exercise-6.9-eq1} holds.
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\paragraph{(ii)}%
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Let $x \in \ran{\bigcap{\mathscr{A}}}$.
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By definition of the range of a set,
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$\exists t, \left< t, x \right> \in \bigcap{\mathscr{A}}$.
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By definition of the intersection of sets,
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$\exists t, \forall R \in \mathscr{A}, \left< t, x \right> \in R$.
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But this implies that
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$\forall R \in \mathscr{A}, \exists t, \left< t, x \right> \in R$.
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By another application of the definition of the domain of a set,
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$\forall R \in \mathscr{A}, x \in \ran{R}$.
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By another application of the intersection of sets,
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$x \in \bigcap\;\{ \ran{R} \mid R \in \mathscr{A} \}$.
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Thus \eqref{sub:exercise-6.9-eq2} holds.
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\end{answer}
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\section{Exercises 7}%
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\label{sec:exercises-7}
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@ -329,4 +329,107 @@ theorem exercise_6_7 {R : Set.Relation α}
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simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this
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exact hxy_mem this
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/-- ### Exercise 6.8i
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Show that for any set `𝓐`:
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```
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dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 }
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```
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-/
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theorem exercise_6_8_i {A : Set (Set.Relation α)}
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: Set.Relation.dom (⋃₀ A) = ⋃₀ { Set.Relation.dom R | R ∈ A } := by
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ext x
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unfold Set.Relation.dom Prod.fst
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simp only [
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Set.mem_image,
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Set.mem_sUnion,
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Prod.exists,
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exists_and_right,
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exists_eq_right,
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Set.mem_setOf_eq,
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exists_exists_and_eq_and
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]
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apply Iff.intro
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· intro ⟨y, ⟨t, ⟨ht, hx⟩⟩⟩
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exact ⟨t, ⟨ht, ⟨y, hx⟩⟩⟩
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· intro ⟨t, ⟨ht, ⟨y, hx⟩⟩⟩
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exact ⟨y, ⟨t, ⟨ht, hx⟩⟩⟩
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/-- ### Exercise 6.8ii
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Show that for any set `𝓐`:
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```
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ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 }
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```
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-/
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theorem exercise_6_8_ii {A : Set (Set.Relation α)}
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: Set.Relation.ran (⋃₀ A) = ⋃₀ { Set.Relation.ran R | R ∈ A } := by
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ext x
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unfold Set.Relation.ran Prod.snd
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simp only [
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Set.mem_image,
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Set.mem_sUnion,
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Prod.exists,
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exists_eq_right,
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Set.mem_setOf_eq,
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exists_exists_and_eq_and
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]
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apply Iff.intro
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· intro ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
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exact ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
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· intro ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
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exact ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
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/-- ## Exercise 6.9i
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Discuss the result of replacing the union operation by the intersection
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operation in the preceding problem.
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```
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dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 }
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```
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-/
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theorem exercise_6_9_i {A : Set (Set.Relation α)}
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: Set.Relation.dom (⋂₀ A) ⊆ ⋂₀ { Set.Relation.dom R | R ∈ A } := by
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show ∀ x, x ∈ Set.Relation.dom (⋂₀ A) → x ∈ ⋂₀ { Set.Relation.dom R | R ∈ A }
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unfold Set.Relation.dom Prod.fst
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simp only [
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Set.mem_image,
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Set.mem_sInter,
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Prod.exists,
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exists_and_right,
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exists_eq_right,
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Set.mem_setOf_eq,
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forall_exists_index,
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and_imp,
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forall_apply_eq_imp_iff₂
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]
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intro _ y hy R hR
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exact ⟨y, hy R hR⟩
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/-- ## Exercise 6.9ii
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Discuss the result of replacing the union operation by the intersection
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operation in the preceding problem.
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```
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ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 }
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```
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-/
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theorem exercise_6_9_ii {A : Set (Set.Relation α)}
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: Set.Relation.ran (⋂₀ A) ⊆ ⋂₀ { Set.Relation.ran R | R ∈ A } := by
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show ∀ x, x ∈ Set.Relation.ran (⋂₀ A) → x ∈ ⋂₀ { Set.Relation.ran R | R ∈ A }
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unfold Set.Relation.ran Prod.snd
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simp only [
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Set.mem_image,
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Set.mem_sInter,
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Prod.exists,
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exists_and_right,
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exists_eq_right,
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Set.mem_setOf_eq,
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forall_exists_index,
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and_imp,
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forall_apply_eq_imp_iff₂
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]
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intro _ y hy R hR
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exact ⟨y, hy R hR⟩
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end Enderton.Set.Chapter_3
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