∪F and ∪G are disjoint iff for all A in F and B in G, A and B are disjoint (Proof)

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(∪F) \ (∪G) = ∪(F \ G) iff for all A in F\G and B in G, A and B are disjoint (Proof) (∪F) \ (∪G) = ∪(F \ G) iff for all A in F\G and B in G, A and B are disjoint (Proof)
∪(F ∩ G) = (∪F) ∩ (∪G) iff for all A in F and B in G, A ∩ B ⊆ ∪(F ∩ G) Proof [ILIEKMATHPHYSICS] ∪(F ∩ G) = (∪F) ∩ (∪G) iff for all A in F and B in G, A ∩ B ⊆ ∪(F ∩ G) Proof [ILIEKMATHPHYSICS]
If ∪F ⊈ ∪G, then there is some A in F such that for all B in G, A ⊈ B (Proof) [ILIEKMATHPHYSICS] If ∪F ⊈ ∪G, then there is some A in F such that for all B in G, A ⊈ B (Proof) [ILIEKMATHPHYSICS]
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