Theorem riinrab 5008

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 5006	. . 3 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = 𝐴)
2		rzal 4455	. . . . 5 ⊢ (𝑋 = ∅ → ∀𝑥 ∈ 𝑋 𝜑)
3	2	ralrimivw 3185	. . . 4 ⊢ (𝑋 = ∅ → ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
4		rabid2 3383	. . . 4 ⊢ (𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑} ↔ ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
5	3, 4	sylibr 236	. . 3 ⊢ (𝑋 = ∅ → 𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
6	1, 5	eqtrd 2858	. 2 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
7		ssrab2 4058	. . . . 5 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
8	7	rgenw 3152	. . . 4 ⊢ ∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
9		riinn0 5007	. . . 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 4993	. . 3 ⊢ (𝑋 ≠ ∅ → ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
12	10, 11	eqtrd 2858	. 2 ⊢ (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
13	6, 12	pm2.61ine 3102	1 ⊢ (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑}

Syntax hints:   = wceq 1537   ≠ wne 3018  ∀wral 3140  {crab 3144   ∩ cin 3937   ⊆ wss 3938  ∅c0 4293  ∩ ciin 4922

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-ne 3019  df-ral 3145  df-rex 3146  df-rab 3149  df-v 3498  df-dif 3941  df-in 3945  df-ss 3954  df-nul 4294  df-iin 4924

This theorem is referenced by:  acsfn1  16934  acsfn1c  16935  acsfn2  16936  cntziinsn  18467  acsfn1p  19580  csscld  23854