Theorem iinrabm 3875

Description: Indexed intersection of a restricted class builder. (Contributed by Jim Kingdon, 16-Aug-2018.)

Assertion

Ref	Expression
iinrabm	⊢ (∃𝑥 𝑥 ∈ 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 iinrabm

Step	Hyp	Ref	Expression
1		r19.28mv 3455	. . 3 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑) ↔ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)))
2	1	abbidv 2257	. 2 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)})
3		df-rab 2425	. . . . 5 ⊢ {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)}
4	3	a1i 9	. . . 4 ⊢ (𝑥 ∈ 𝐴 → {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)})
5	4	iineq2i 3832	. . 3 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = ∩ 𝑥 ∈ 𝐴 {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)}
6		iinab 3874	. . 3 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)} = {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)}
7	5, 6	eqtri 2160	. 2 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)}
8		df-rab 2425	. 2 ⊢ {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)}
9	2, 7, 8	3eqtr4g 2197	1 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑})

Syntax hints:   → wi 4   ∧ wa 103   = wceq 1331  ∃wex 1468   ∈ wcel 1480  {cab 2125  ∀wral 2416  {crab 2420  ∩ ciin 3814

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-ral 2421  df-rab 2425  df-v 2688  df-iin 3816

This theorem is referenced by: (None)