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Theorem iinrab2 4581
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 4533 . . . . . 6 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = 𝑥 ∈ ∅ {𝑦𝐵𝜑})
2 0iin 4576 . . . . . 6 𝑥 ∈ ∅ {𝑦𝐵𝜑} = V
31, 2syl6eq 2671 . . . . 5 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = V)
43ineq1d 3811 . . . 4 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = (V ∩ 𝐵))
5 incom 3803 . . . . 5 (V ∩ 𝐵) = (𝐵 ∩ V)
6 inv1 3968 . . . . 5 (𝐵 ∩ V) = 𝐵
75, 6eqtri 2643 . . . 4 (V ∩ 𝐵) = 𝐵
84, 7syl6eq 2671 . . 3 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = 𝐵)
9 rzal 4071 . . . 4 (𝐴 = ∅ → ∀𝑥𝐴𝑦𝐵 𝜑)
10 rabid2 3116 . . . . 5 (𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ↔ ∀𝑦𝐵𝑥𝐴 𝜑)
11 ralcom 3096 . . . . 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 2655 . 2 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
15 iinrab 4580 . . . 4 (𝐴 ≠ ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
1615ineq1d 3811 . . 3 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵))
17 ssrab2 3685 . . . 4 {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ⊆ 𝐵
18 dfss 3587 . . . 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 2673 . 2 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
2114, 20pm2.61ine 2876 1 ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑}
Colors of variables: wff setvar class
Syntax hints:   = wceq 1482  wne 2793  wral 2911  {crab 2915  Vcvv 3198  cin 3571  wss 3572  c0 3913   ciin 4519
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1721  ax-4 1736  ax-5 1838  ax-6 1887  ax-7 1934  ax-9 1998  ax-10 2018  ax-11 2033  ax-12 2046  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1485  df-ex 1704  df-nf 1709  df-sb 1880  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2752  df-ne 2794  df-ral 2916  df-rab 2920  df-v 3200  df-dif 3575  df-in 3579  df-ss 3586  df-nul 3914  df-iin 4521
This theorem is referenced by: (None)
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