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Theorem riinrab 5047
Description: Relative intersection of a relative abstraction. (Contributed by Stefan O'Rear, 3-Apr-2015.)
Assertion
Ref Expression
riinrab (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑}
Distinct variable groups:   𝑥,𝐴,𝑦   𝑥,𝑋,𝑦
Allowed substitution hints:   𝜑(𝑥, 𝑦)

Proof of Theorem riinrab
StepHypRef Expression
1 riin0 5045 . . 3 (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = 𝐴)
2 rzal 4453 . . . . 5 (𝑋 = ∅ → ∀𝑥 ∈ 𝑋 𝜑)
32ralrimivw 3162 . . . 4 (𝑋 = ∅ → ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
4 rabid2 3448 . . . 4 (𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑} ↔ ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
53, 4sylibr 237 . . 3 (𝑋 = ∅ → 𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
61, 5eqtrd 2798 . 2 (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
7 ssrab2 4008 . . . . 5 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
87rgenw 3084 . . . 4 ∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
9 riinn0 5046 . . . 4 ((∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴 ∧ 𝑋 ≠ ∅) → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑})
108, 9mpan 703 . . 3 (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑})
11 iinrab 5030 . . 3 (𝑋 ≠ ∅ → ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
1210, 11eqtrd 2798 . 2 (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
136, 12pm2.61ine 3042 1 (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑}
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   = wceq 1570   ≠ wne 2959  ∀wral 3080  {crab 3416   ⊆ wss 3900   ∩ cin 4092  ∅c0 4281  ∩ ciin 4955
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-nul 4282  df-iin 4957
This theorem is used by:  acsfn1  17861  acsfn1c  17862  acsfn2  17863  cntziinsn  19578  acsfn1p  21083  csscld  25597
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