May someone perform a quick check of this proof involving disjunctions?

[Syrus]

Homework Statement

*below, the notation P() denotes the power set of the set within parentheses.

Prove that for any sets A and B, if P(A) U P(B) = P(A U B) then either A ⊆ B or B ⊆ A.

Homework Equations

The Attempt at a Solution

Let A and B be arbitrary. Assume P(A) U P(B) = P(A U B). Note that A U B ⊆ A U B (trivial lemma omitted). This means A U B ∈ P(A U B). By our original assumption, then, A U B ∈ P(A) U P(B). So either A U B ∈ P(A) or A U B ∈ P(B).
Case I. A U B ∈ P(A). Then A U B ⊆ A. Now let x ∈ B. It clearly follows that x ∈ A or x ∈ B. Hence, x ∈ A U B. But since A U B ⊆ A, x ∈ A. As x was arbitrary, B ⊆ A. Obviously, this implies that either A ⊆ B or B ⊆ A.
Case II. A U B ∈ P(B). Then A U B ⊆ B. Now let x ∈ A. The proof of this case is analogous to that in Case I.
Since these cases are exhaustive and each result in A ⊆ B or B ⊆ A, if P(A) U P(B) = P(A U B) then either A ⊆ B or B ⊆ A. Lastly, since A and B were arbitrary, then we have shown for any sets A and B, if P(A) U P(B) = P(A U B) then either A ⊆ B or B ⊆ A.

 

[Dick]

Homework Statement

*below, the notation P() denotes the power set of the set within parentheses.

Prove that for any sets A and B, if P(A) U P(B) = P(A U B) then either A ⊆ B or B ⊆ A.

Homework Equations

The Attempt at a Solution

Let A and B be arbitrary. Assume P(A) U P(B) = P(A U B). Note that A U B ⊆ A U B (trivial lemma omitted). This means A U B ∈ P(A U B). By our original assumption, then, A U B ∈ P(A) U P(B). So either A U B ∈ P(A) or A U B ∈ P(B).
Case I. A U B ∈ P(A). Then A U B ⊆ A. Now let x ∈ B. It clearly follows that x ∈ A or x ∈ B. Hence, x ∈ A U B. But since A U B ⊆ A, x ∈ A. As x was arbitrary, B ⊆ A. Obviously, this implies that either A ⊆ B or B ⊆ A.
Case II. A U B ∈ P(B). Then A U B ⊆ B. Now let x ∈ A. The proof of this case is analogous to that in Case I.
Since these cases are exhaustive and each result in A ⊆ B or B ⊆ A, if P(A) U P(B) = P(A U B) then either A ⊆ B or B ⊆ A. Lastly, since A and B were arbitrary, then we have shown for any sets A and B, if P(A) U P(B) = P(A U B) then either A ⊆ B or B ⊆ A.

Looks fine to me.