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3:27 AM
The easiest way to prove the required sentence is to expand the proof for ∀S∈nset ∃I∈set ∀x∈obj ( x∈I ⇔ ∀T∈set ( T∈I ⇒ x∈T ) ):
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3:53 AM
> ∀S∈nset ∀I∈set ∀x∈obj ( x∈Intersect(S) ⇔ ∀T∈set ( T∈S ⇒ x∈T ) ).
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> ∀S∈set ∀x∈obj ( x∈Union(S) ⇔ ∃∈set ( T∈S ⇒ x∈T ) ).
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