Prove that for all sets A and B, A ∩ B = ∅ implies ( A ∪ B ) - B = A

Question

In the next proof we avail ourselves of the next lemma:

For all sets A and B, ( A ∪ B ) - B = A.

Proof:

Let A and B be arbitrary sets and let x ∈ ( A ∪ B ) - B.

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

For p ∧ ~ p ≡ F and p ∧ q ⇒ p where p and q are prepositions,

( x ∈ A ∧ x ∉ B ) ∨ ( x ∈ B ∨ x ∉ B ) ⇔ x ∈ A ∨ F

By modus tollendo ponens,

x ∈ A ∨ F ⇔ x ∈ A

So then,

x ∈ ( A ∪ B ) - B ⇔ x ∈ A

And

( A ∪ B ) - B = A

Therefore, the statement for all sets A and B, A ∩ B = ∅ implies ( A ∪ B ) - B = A is true trivially.

Is this proof right?

Answer 1

Consider $$B \subset A, B \ne \emptyset, A \setminus B = C \ne \emptyset$$ Then, because B is non empty $$A \cup B = A \ne A \setminus B$$

Answer 2

$A\cap B=\varnothing\implies A\subset \bar B\implies A\cap \bar B=A$

Thus $(A\cup B)-B=(A\cup B)\cap\bar B=(A\cap\bar B)\cup(B\cap\bar B)=A\cup\varnothing=A$