Theorem cover2g 36887

Description: Two ways of expressing the statement "there is a cover of 𝐴 by elements of 𝐵 such that for each set in the cover, 𝜑". Note that 𝜑 and 𝑥 must be distinct. (Contributed by Jeff Madsen, 21-Jun-2010.)

Hypothesis

Ref	Expression
cover2g.1	⊢ 𝐴 = ∪ 𝐵

Assertion

Ref	Expression
cover2g	⊢ (𝐵 ∈ 𝐶 → (∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 ∈ 𝑦 ∧ 𝜑) ↔ ∃𝑧 ∈ 𝒫 𝐵(∪ 𝑧 = 𝐴 ∧ ∀𝑦 ∈ 𝑧 𝜑)))

Distinct variable groups:   𝜑,𝑥,𝑧   𝑥,𝐵,𝑦,𝑧   𝑥,𝐴,𝑧

Allowed substitution hints:   𝜑(𝑦)   𝐴(𝑦)   𝐶(𝑥,𝑦,𝑧)

Proof of Theorem cover2g

Dummy variable 𝑏 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		unieq 4918	. . . 4 ⊢ (𝑏 = 𝐵 → ∪ 𝑏 = ∪ 𝐵)
2		cover2g.1	. . . 4 ⊢ 𝐴 = ∪ 𝐵
3	1, 2	eqtr4di 2788	. . 3 ⊢ (𝑏 = 𝐵 → ∪ 𝑏 = 𝐴)
4		rexeq 3319	. . 3 ⊢ (𝑏 = 𝐵 → (∃𝑦 ∈ 𝑏 (𝑥 ∈ 𝑦 ∧ 𝜑) ↔ ∃𝑦 ∈ 𝐵 (𝑥 ∈ 𝑦 ∧ 𝜑)))
5	3, 4	raleqbidv 3340	. 2 ⊢ (𝑏 = 𝐵 → (∀𝑥 ∈ ∪ 𝑏∃𝑦 ∈ 𝑏 (𝑥 ∈ 𝑦 ∧ 𝜑) ↔ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 ∈ 𝑦 ∧ 𝜑)))
6		pweq 4615	. . 3 ⊢ (𝑏 = 𝐵 → 𝒫 𝑏 = 𝒫 𝐵)
7	3	eqeq2d 2741	. . . 4 ⊢ (𝑏 = 𝐵 → (∪ 𝑧 = ∪ 𝑏 ↔ ∪ 𝑧 = 𝐴))
8	7	anbi1d 628	. . 3 ⊢ (𝑏 = 𝐵 → ((∪ 𝑧 = ∪ 𝑏 ∧ ∀𝑦 ∈ 𝑧 𝜑) ↔ (∪ 𝑧 = 𝐴 ∧ ∀𝑦 ∈ 𝑧 𝜑)))
9	6, 8	rexeqbidv 3341	. 2 ⊢ (𝑏 = 𝐵 → (∃𝑧 ∈ 𝒫 𝑏(∪ 𝑧 = ∪ 𝑏 ∧ ∀𝑦 ∈ 𝑧 𝜑) ↔ ∃𝑧 ∈ 𝒫 𝐵(∪ 𝑧 = 𝐴 ∧ ∀𝑦 ∈ 𝑧 𝜑)))
10		vex 3476	. . 3 ⊢ 𝑏 ∈ V
11		eqid 2730	. . 3 ⊢ ∪ 𝑏 = ∪ 𝑏
12	10, 11	cover2 36886	. 2 ⊢ (∀𝑥 ∈ ∪ 𝑏∃𝑦 ∈ 𝑏 (𝑥 ∈ 𝑦 ∧ 𝜑) ↔ ∃𝑧 ∈ 𝒫 𝑏(∪ 𝑧 = ∪ 𝑏 ∧ ∀𝑦 ∈ 𝑧 𝜑))
13	5, 9, 12	vtoclbg 3543	1 ⊢ (𝐵 ∈ 𝐶 → (∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 ∈ 𝑦 ∧ 𝜑) ↔ ∃𝑧 ∈ 𝒫 𝐵(∪ 𝑧 = 𝐴 ∧ ∀𝑦 ∈ 𝑧 𝜑)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 394   = wceq 1539   ∈ wcel 2104  ∀wral 3059  ∃wrex 3068  𝒫 cpw 4601  ∪ cuni 4907

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-10 2135  ax-11 2152  ax-12 2169  ax-ext 2701  ax-sep 5298

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-tru 1542  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-nfc 2883  df-ral 3060  df-rex 3069  df-rab 3431  df-v 3474  df-in 3954  df-ss 3964  df-pw 4603  df-uni 4908

This theorem is referenced by: (None)