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Theorem iinrab2 5035
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 4975 . . . . . 6 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = 𝑥 ∈ ∅ {𝑦𝐵𝜑})
2 0iin 5029 . . . . . 6 𝑥 ∈ ∅ {𝑦𝐵𝜑} = V
31, 2eqtrdi 2820 . . . . 5 (𝐴 = ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = V)
43ineq1d 4180 . . . 4 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = (V ∩ 𝐵))
5 inv1 4361 . . . . 5 (𝐵 ∩ V) = 𝐵
65ineqcomi 4172 . . . 4 (V ∩ 𝐵) = 𝐵
74, 6eqtrdi 2820 . . 3 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = 𝐵)
8 rzal 4457 . . . 4 (𝐴 = ∅ → ∀𝑥𝐴𝑦𝐵 𝜑)
9 rabid2 3456 . . . . 5 (𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ↔ ∀𝑦𝐵𝑥𝐴 𝜑)
10 ralcom 3299 . . . . 5 (∀𝑦𝐵𝑥𝐴 𝜑 ↔ ∀𝑥𝐴𝑦𝐵 𝜑)
119, 10bitr2i 279 . . . 4 (∀𝑥𝐴𝑦𝐵 𝜑𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
128, 11sylib 221 . . 3 (𝐴 = ∅ → 𝐵 = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
137, 12eqtrd 2804 . 2 (𝐴 = ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
14 iinrab 5034 . . . 4 (𝐴 ≠ ∅ → 𝑥𝐴 {𝑦𝐵𝜑} = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
1514ineq1d 4180 . . 3 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵))
16 ssrab2 4042 . . . 4 {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ⊆ 𝐵
17 dfss 3932 . . . 4 ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ⊆ 𝐵 ↔ {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵))
1816, 17mpbi 233 . . 3 {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} = ({𝑦𝐵 ∣ ∀𝑥𝐴 𝜑} ∩ 𝐵)
1915, 18eqtr4di 2822 . 2 (𝐴 ≠ ∅ → ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑})
2013, 19pm2.61ine 3047 1 ( 𝑥𝐴 {𝑦𝐵𝜑} ∩ 𝐵) = {𝑦𝐵 ∣ ∀𝑥𝐴 𝜑}
Colors of variables: wff setvar class
Syntax hints:   = wceq 1567  wne 2964  wral 3085  {crab 3423  Vcvv 3463  cin 3912  wss 3913  c0 4294   ciin 4958
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-10 2182  ax-11 2198  ax-12 2219  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-nf 1811  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-nfc 2918  df-ne 2965  df-ral 3086  df-rex 3096  df-rab 3424  df-v 3465  df-dif 3916  df-in 3920  df-ss 3930  df-nul 4295  df-iin 4960
This theorem is referenced by: (None)
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