Theorem iinrabm 4038

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 3589	. . 3 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑) ↔ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)))
2	1	abbidv 2350	. 2 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)})
3		df-rab 2520	. . . . 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 3994	. . 3 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = ∩ 𝑥 ∈ 𝐴 {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)}
6		iinab 4037	. . 3 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜑)} = {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)}
7	5, 6	eqtri 2252	. 2 ⊢ ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∣ ∀𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∧ 𝜑)}
8		df-rab 2520	. 2 ⊢ {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜑)}
9	2, 7, 8	3eqtr4g 2289	1 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑})

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1398  ∃wex 1541   ∈ wcel 2202  {cab 2217  ∀wral 2511  {crab 2515  ∩ ciin 3976

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-ral 2516  df-rab 2520  df-v 2805  df-iin 3978

This theorem is referenced by: (None)