text‹
Zorn's Lemmas states: if every linear ordered subset of an ordered set ‹S›
has an upper bound in ‹S›, then there exists a maximal element in ‹S›. In
our application, ‹S› is a set of sets ordered by set inclusion. Since the
union of a chain of sets is an upper bound for all elements of the chain,
the conditions of Zorn's lemma can be modified: if ‹S› is non-empty, it
suffices to show that for every non-empty chain ‹c› in ‹S› the union of ‹c›
also lies in ‹S›. ›
theorem Zorn's_Lemma: assumes r: "∧c. c ∈ chains S ==>∃x. x ∈ c ==>∪c ∈ S" and aS: "a ∈ S" shows"∃y ∈ S. ∀z ∈ S. y ⊆ z ⟶ z = y" proof (rule Zorn_Lemma2) show"∀c ∈ chains S. ∃y ∈ S. ∀z ∈ c. z ⊆ y" proof fix c assume"c ∈ chains S" show"∃y ∈ S. ∀z ∈ c. z ⊆ y" proof (cases "c = {}") txt‹If ‹c› is an empty chain, then every element in ‹S› is an upper
bound of ‹c›.› case True with aS show ?thesis by fast next txt‹If ‹c› is non-empty, then ‹∪c› is an upper bound of ‹c›, lying in ‹S›.› case False show ?thesis proof show"∀z ∈ c. z ⊆∪c"by fast show"∪c ∈ S" proof (rule r) from‹c ≠ {}›show"∃x. x ∈ c"by fast show"c ∈ chains S"by fact qed qed qed qed qed
end
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