Assume that A ⊆ B, then x ∈ A ⇒ x ∈ B.

Theorem 1. (p ⇒ q) ⇒ [(p ∨ r) ⇒ (q ∨ r)]
Theorem 2. p ∨ p ⇔ p.
Theorem 3. For all classes A and B, A, B ⊆ A ∪ B.

x ∈ A ∪ B
⇒ x ∈ A ∨ x ∈ B
⇒ x ∈ B ∨ x ∈ B by Thm 1.
⇒ x ∈ B by Thm 2.

Thus, A ∪ B ⊆ B, then B ⊆ A ∪ B, by Thm 3.
A ∪ B = B.

Assume that A ∪ B = B, then A ⊆ A ∪ B, by Thm 3.
Thus A ⊆ A ∪ B = B.

Consequently, A ⊆ B ⇔ A ∪ B = B.

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