Prove that for A, B and C sets, A - ( B - C ) = ( A - B ) ∪ ( A ∩ C ) (Alternative to the proof given)

Question

My proof:

Let A, B and C be arbitrary sets.

As a means to prove such a statement we are going to verify that

x ∈ A - ( B - C ) ⇔ x ∈ ( A - B ) ∪ ( A ∩ C )

Note that

x ∈ A - ( B - C ) ⇔ x ∈ A ∧ x ∉ ( B - C ) ⇔ x ∈ A ∧ ( x ∉ B ∨ x ∈ C ) ⇔ ( x ∈ A ∧ x ∉ B ) ∨ ( x ∈ A ∧ x ∈ C ) ⇔ x ∈ ( A - B ) ∨ x ∈ ( A ∩ C ) ⇔ x ∈ ( A - B ) ∪ ( A ∩ C )

Hence,

A - ( B - C ) = ( A - B ) ∪ ( A ∩ C )

Answer 1

$$A\setminus(B\setminus C) = A \cap (B\setminus C)^C $$ $$=A\cap(B\cap C^c)^c = A\cap(B^c \cup C)$$ $$= (A\cap B^c)\cup(A\cap C)$$ $$=(A\setminus B)\cup(A\cap C)$$