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Theorem iinrab2 4615
Description: Indexed intersection of a restricted class builder. (Contributed by NM, 6-Dec-2011.)
Assertion
Ref Expression
iinrab2 ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑}
Distinct variable groups:   𝑦,𝐴,𝑥   𝑥,𝐵,𝑦
Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem iinrab2
StepHypRef Expression
1 iineq1 4567 . . . . . 6 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = 𝑥 ∈ ∅ {𝑦𝐵𝜑})
2 0iin 4610 . . . . . 6 𝑥 ∈ ∅ {𝑦𝐵𝜑} = V
31, 2syl6eq 2701 . . . . 5 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = V)
43ineq1d 3846 . . . 4 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = (V ∩ 𝐵))
5 incom 3838 . . . . 5 (V ∩ 𝐵) = (𝐵 ∩ V)
6 inv1 4003 . . . . 5 (𝐵 ∩ V) = 𝐵
75, 6eqtri 2673 . . . 4 (V ∩ 𝐵) = 𝐵
84, 7syl6eq 2701 . . 3 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = 𝐵)
9 rzal 4106 . . . 4 (𝐴 = ∅ → ∀𝑥𝐴𝑦𝐵 𝜑)
10 rabid2 3148 . . . . 5 (𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ↔ ∀𝑦𝐵𝑥𝐴 𝜑)
11 ralcom 3127 . . . . 5 (∀𝑦𝐵𝑥𝐴 𝜑 ↔ ∀𝑥𝐴𝑦𝐵 𝜑)
1210, 11bitr2i 265 . . . 4 (∀𝑥𝐴𝑦𝐵 𝜑𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
139, 12sylib 208 . . 3 (𝐴 = ∅ → 𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
148, 13eqtrd 2685 . 2 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
15 iinrab 4614 . . . 4 (𝐴 ≠ ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
1615ineq1d 3846 . . 3 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵))
17 ssrab2 3720 . . . 4 {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ⊆ 𝐵
18 dfss 3622 . . . 4 ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ⊆ 𝐵 ↔ {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵))
1917, 18mpbi 220 . . 3 {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵)
2016, 19syl6eqr 2703 . 2 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
2114, 20pm2.61ine 2906 1 ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑}
Colors of variables: wff setvar class
Syntax hints:   = wceq 1523  wne 2823  wral 2941  {crab 2945  Vcvv 3231  cin 3606  wss 3607  c0 3948   ciin 4553
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-nfc 2782  df-ne 2824  df-ral 2946  df-rab 2950  df-v 3233  df-dif 3610  df-in 3614  df-ss 3621  df-nul 3949  df-iin 4555
This theorem is referenced by: (None)
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