Prove $(A∩B)’=A’∪B’$.
Let $x ∈ (A ∩ B)’$
$∴ (x ∈ A ∩ B)’$
$∴ (x ∈ A ∧ x ∈ B)’$
$∴ (x ∈ A)’ ∨ (x ∈ B)’$
$∴ x ∈ A’ ∨ x ∈ B’$
$∴ x ∈ A’ ∪ B’$
$∴ (A ∩ B)’ ⊆ A’ ∪ B’$
The above is the solution provided. Sorry if it seems trivial but I don't understand how
$∴ (x ∈ A ∧ x ∈ B)’$
leads to
$∴ (x ∈ A)’ ∨ (x ∈ B)’$
and not
$∴ (x ∈ A)’ ∧ (x ∈ B)’$