Theorem iinrab 5023

Description: Indexed intersection of a restricted class abstraction. (Contributed by NM, 6-Dec-2011.)

Assertion

Ref	Expression
iinrab	⊢ (𝐴 ≠ ∅ → ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑})

Distinct variable groups:   𝑦,𝐴,𝑥   𝑥,𝐵

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝐵(𝑦)

Proof of Theorem iinrab

Step	Hyp	Ref	Expression
1		r19.28zv 4457	. . 3 ⊢ (𝐴 ≠ ∅ → (∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑) ↔ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)))
2	1	abbidv 2827	. 2 ⊢ (𝐴 ≠ ∅ → {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)})
3		df-rab 3414	. . . . 5 ⊢ {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)}
4	3	a1i 11	. . . 4 ⊢ (𝑥 ∈ 𝐴 → {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)})
5	4	iineq2i 4969	. . 3 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = ∩ 𝑥 ∈ 𝐴 {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)}
6		iinab 5022	. . 3 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)} = {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)}
7	5, 6	eqtri 2784	. 2 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)}
8		df-rab 3414	. 2 ⊢ {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)}
9	2, 7, 8	3eqtr4g 2821	1 ⊢ (𝐴 ≠ ∅ → ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑})

Syntax hints:   → wi 4   ∧ wa 399   = wceq 1559   ∈ wcel 2141  {cab 2739   ≠ wne 2956  ∀wral 3075  {crab 3413  ∅c0 4283  ∩ ciin 4947

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-fal 1572  df-ex 1799  df-nf 1803  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ne 2957  df-ral 3076  df-rab 3414  df-v 3455  df-dif 3905  df-nul 4284  df-iin 4949

This theorem is referenced by:  iinrab2  5024  riinrab  5038  ubthlem1  31030  pmapglbx  40354  preimageiingt  47255  preimaleiinlt  47256