Theorem 1
THEOREM 1 If A and B are any two sets,then
$(i) A - B = A \cap B' (ii) B - A = B \cap A' (iii) A - B = A \Leftrightarrow A \cap B = ϕ (iv) (A - B) \cup B = A \cup B (v) (A - B) \cap B = ϕ (vi) A \subseteq B \Leftrightarrow B' \subseteq A' (vii) (A - B) \cup (B - A) = (A \cup B) - (A \cap B)$

Question

Theorem 1
THEOREM 1 If A and B are any two sets,then
i   A - B = A ∩ B ' ( ii )   B - A = B ∩ A ' (
iii )   A - B = A ⇔ A ∩ B = ϕ ( iv )   ( A
- B ) ∪ B = A ∪ B ( v )   A - B ∩ B = ϕ ( vi
)   A ⊆ B ⇔ B ' ⊆ A ' ( vii )   ( A - B )
∪ ( B - A ) = ( A ∪ B ) - ( A ∩ B )

Sanjari Kalantri · Accepted Answer

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Theorem 1 THEOREM 1 If A and B are any two sets,then i A - B = A ∩ B ' ( ii ) B - A = B ∩ A ' ( iii ) A - B = A ⇔ A ∩ B = ϕ ( iv ) ( A - B ) ∪ B = A ∪ B ( v ) A - B ∩ B = ϕ ( vi ) A ⊆ B ⇔ B ' ⊆ A ' ( vii ) ( A - B ) ∪ ( B - A ) = ( A ∪ B ) - ( A ∩ B )