Theorem riinrab 5079

Description: Relative intersection of a relative abstraction. (Contributed by Stefan O'Rear, 3-Apr-2015.)

Assertion

Ref	Expression
riinrab	⊢ (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑}

Distinct variable groups:   𝑥,𝐴,𝑦   𝑥,𝑋,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem riinrab

Step	Hyp	Ref	Expression
1		riin0 5077	. . 3 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = 𝐴)
2		rzal 4501	. . . . 5 ⊢ (𝑋 = ∅ → ∀𝑥 ∈ 𝑋 𝜑)
3	2	ralrimivw 3149	. . . 4 ⊢ (𝑋 = ∅ → ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
4		rabid2 3463	. . . 4 ⊢ (𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑} ↔ ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
5	3, 4	sylibr 233	. . 3 ⊢ (𝑋 = ∅ → 𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
6	1, 5	eqtrd 2771	. 2 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
7		ssrab2 4072	. . . . 5 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
8	7	rgenw 3064	. . . 4 ⊢ ∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
9		riinn0 5078	. . . 4 ⊢ ((∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴 ∧ 𝑋 ≠ ∅) → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑})
10	8, 9	mpan 688	. . 3 ⊢ (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑})
11		iinrab 5064	. . 3 ⊢ (𝑋 ≠ ∅ → ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
12	10, 11	eqtrd 2771	. 2 ⊢ (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
13	6, 12	pm2.61ine 3024	1 ⊢ (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑}

Syntax hints:   = wceq 1541   ≠ wne 2939  ∀wral 3060  {crab 3431   ∩ cin 3942   ⊆ wss 3943  ∅c0 4317  ∩ ciin 4990

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-fal 1554  df-ex 1782  df-nf 1786  df-sb 2068  df-clab 2709  df-cleq 2723  df-clel 2809  df-nfc 2884  df-ne 2940  df-ral 3061  df-rex 3070  df-rab 3432  df-v 3474  df-dif 3946  df-in 3950  df-ss 3960  df-nul 4318  df-iin 4992

This theorem is referenced by:  acsfn1  17586  acsfn1c  17587  acsfn2  17588  cntziinsn  19164  acsfn1p  20361  csscld  24692