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Theorem iinrab2 5012
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 4951 . . . . . 6 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = 𝑥 ∈ ∅ {𝑦𝐵𝜑})
2 0iin 5006 . . . . . 6 𝑥 ∈ ∅ {𝑦𝐵𝜑} = V
31, 2eqtrdi 2787 . . . . 5 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = V)
43ineq1d 4159 . . . 4 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = (V ∩ 𝐵))
5 inv1 4338 . . . . 5 (𝐵 ∩ V) = 𝐵
65ineqcomi 4151 . . . 4 (V ∩ 𝐵) = 𝐵
74, 6eqtrdi 2787 . . 3 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = 𝐵)
8 rzal 4434 . . . 4 (𝐴 = ∅ → ∀𝑥𝐴𝑦𝐵 𝜑)
9 rabid2 3422 . . . . 5 (𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ↔ ∀𝑦𝐵𝑥𝐴 𝜑)
10 ralcom 3265 . . . . 5 (∀𝑦𝐵𝑥𝐴 𝜑 ↔ ∀𝑥𝐴𝑦𝐵 𝜑)
119, 10bitr2i 276 . . . 4 (∀𝑥𝐴𝑦𝐵 𝜑𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
128, 11sylib 218 . . 3 (𝐴 = ∅ → 𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
137, 12eqtrd 2771 . 2 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
14 iinrab 5011 . . . 4 (𝐴 ≠ ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
1514ineq1d 4159 . . 3 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵))
16 ssrab2 4020 . . . 4 {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ⊆ 𝐵
17 dfss 3908 . . . 4 ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ⊆ 𝐵 ↔ {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵))
1816, 17mpbi 230 . . 3 {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵)
1915, 18eqtr4di 2789 . 2 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
2013, 19pm2.61ine 3015 1 ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑}
Colors of variables: wff setvar class
Syntax hints:   = wceq 1542  wne 2932  wral 3051  {crab 3389  Vcvv 3429  cin 3888  wss 3889  c0 4273   ciin 4934
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 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-nfc 2885  df-ne 2933  df-ral 3052  df-rex 3062  df-rab 3390  df-v 3431  df-dif 3892  df-in 3896  df-ss 3906  df-nul 4274  df-iin 4936
This theorem is referenced by: (None)
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