Theorem iinrabm 3883

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 3460	. . 3 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑) ↔ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)))
2	1	abbidv 2258	. 2 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)})
3		df-rab 2426	. . . . 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 3840	. . 3 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = ∩ 𝑥 ∈ 𝐴 {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)}
6		iinab 3882	. . 3 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)} = {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)}
7	5, 6	eqtri 2161	. 2 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)}
8		df-rab 2426	. 2 ⊢ {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)}
9	2, 7, 8	3eqtr4g 2198	1 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑})

Syntax hints:   → wi 4   ∧ wa 103   = wceq 1332  ∃wex 1469   ∈ wcel 1481  {cab 2126  ∀wral 2417  {crab 2421  ∩ ciin 3822

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 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122

This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-rab 2426  df-v 2691  df-iin 3824

This theorem is referenced by: (None)