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Theorem iinrab2 4999
Description: Indexed intersection of a restricted class abstraction. (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 4939 . . . . . 6 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = 𝑥 ∈ ∅ {𝑦𝐵𝜑})
2 0iin 4993 . . . . . 6 𝑥 ∈ ∅ {𝑦𝐵𝜑} = V
31, 2eqtrdi 2790 . . . . 5 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = V)
43ineq1d 4148 . . . 4 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = (V ∩ 𝐵))
5 inv1 4326 . . . . 5 (𝐵 ∩ V) = 𝐵
65ineqcomi 4140 . . . 4 (V ∩ 𝐵) = 𝐵
74, 6eqtrdi 2790 . . 3 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = 𝐵)
8 rzal 4422 . . . 4 (𝐴 = ∅ → ∀𝑥𝐴𝑦𝐵 𝜑)
9 rabid2 3424 . . . . 5 (𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ↔ ∀𝑦𝐵𝑥𝐴 𝜑)
10 ralcom 3267 . . . . 5 (∀𝑦𝐵𝑥𝐴 𝜑 ↔ ∀𝑥𝐴𝑦𝐵 𝜑)
119, 10bitr2i 277 . . . 4 (∀𝑥𝐴𝑦𝐵 𝜑𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
128, 11sylib 219 . . 3 (𝐴 = ∅ → 𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
137, 12eqtrd 2774 . 2 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
14 iinrab 4998 . . . 4 (𝐴 ≠ ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
1514ineq1d 4148 . . 3 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵))
16 ssrab2 4011 . . . 4 {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ⊆ 𝐵
17 dfss 3902 . . . 4 ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ⊆ 𝐵 ↔ {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵))
1816, 17mpbi 231 . . 3 {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵)
1915, 18eqtr4di 2792 . 2 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
2013, 19pm2.61ine 3017 1 ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑}
Colors of variables: wff setvar class
Syntax hints:   = wceq 1547  wne 2934  wral 3053  {crab 3391  Vcvv 3431  cin 3882  wss 3883  c0 4261   ciin 4922
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-10 2152  ax-11 2168  ax-12 2189  ax-ext 2711
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-nf 1791  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-nfc 2888  df-ne 2935  df-ral 3054  df-rex 3064  df-rab 3392  df-v 3433  df-dif 3886  df-in 3890  df-ss 3900  df-nul 4262  df-iin 4924
This theorem is referenced by: (None)
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