Theorem pwun 5559

Description: The power class of the union of two classes equals the union of their power classes, iff one class is a subclass of the other. Part of Exercise 7(b) of [Enderton] p. 28. (Contributed by NM, 23-Nov-2003.)

Assertion

Ref	Expression
pwun	⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ 𝒫 (𝐴 ∪ 𝐵) = (𝒫 𝐴 ∪ 𝒫 𝐵))

Proof of Theorem pwun

Step	Hyp	Ref	Expression
1		pwunss 4585	. . 3 ⊢ (𝒫 𝐴 ∪ 𝒫 𝐵) ⊆ 𝒫 (𝐴 ∪ 𝐵)
2	1	biantru 539	. 2 ⊢ (𝒫 (𝐴 ∪ 𝐵) ⊆ (𝒫 𝐴 ∪ 𝒫 𝐵) ↔ (𝒫 (𝐴 ∪ 𝐵) ⊆ (𝒫 𝐴 ∪ 𝒫 𝐵) ∧ (𝒫 𝐴 ∪ 𝒫 𝐵) ⊆ 𝒫 (𝐴 ∪ 𝐵)))
3		pwssun 5558	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ 𝒫 (𝐴 ∪ 𝐵) ⊆ (𝒫 𝐴 ∪ 𝒫 𝐵))
4		eqss 3955	. 2 ⊢ (𝒫 (𝐴 ∪ 𝐵) = (𝒫 𝐴 ∪ 𝒫 𝐵) ↔ (𝒫 (𝐴 ∪ 𝐵) ⊆ (𝒫 𝐴 ∪ 𝒫 𝐵) ∧ (𝒫 𝐴 ∪ 𝒫 𝐵) ⊆ 𝒫 (𝐴 ∪ 𝐵)))
5	2, 3, 4	3bitr4i 306	1 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ 𝒫 (𝐴 ∪ 𝐵) = (𝒫 𝐴 ∪ 𝒫 𝐵))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570   ∪ cun 3906   ⊆ wss 3908  𝒫 cpw 4567

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738  ax-sep 5262  ax-pr 5409

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-v 3460  df-un 3913  df-ss 3925  df-pw 4569  df-sn 4595  df-pr 4597

This theorem is used by: (None)