Theorem opabdm 32689

Description: Domain of an ordered-pair class abstraction. (Contributed by Thierry Arnoux, 31-Aug-2017.)

Assertion

Ref	Expression
opabdm	⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → dom 𝑅 = {𝑥 ∣ ∃𝑦𝜑})

Distinct variable group:   𝑥,𝑦,𝑅

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem opabdm

Step	Hyp	Ref	Expression
1		df-dm 5634	. 2 ⊢ dom 𝑅 = {𝑥 ∣ ∃𝑦 𝑥𝑅𝑦}
2		nfopab1 5168	. . . 4 ⊢ Ⅎ𝑥{⟨𝑥, 𝑦⟩ ∣ 𝜑}
3	2	nfeq2 2916	. . 3 ⊢ Ⅎ𝑥 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
4		nfopab2 5169	. . . . 5 ⊢ Ⅎ𝑦{⟨𝑥, 𝑦⟩ ∣ 𝜑}
5	4	nfeq2 2916	. . . 4 ⊢ Ⅎ𝑦 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
6		df-br 5099	. . . . 5 ⊢ (𝑥𝑅𝑦 ↔ ⟨𝑥, 𝑦⟩ ∈ 𝑅)
7		eleq2 2825	. . . . . 6 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ ⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑}))
8		opabidw 5472	. . . . . 6 ⊢ (⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ↔ 𝜑)
9	7, 8	bitrdi 287	. . . . 5 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ 𝜑))
10	6, 9	bitrid 283	. . . 4 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (𝑥𝑅𝑦 ↔ 𝜑))
11	5, 10	exbid 2230	. . 3 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (∃𝑦 𝑥𝑅𝑦 ↔ ∃𝑦𝜑))
12	3, 11	abbid 2804	. 2 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → {𝑥 ∣ ∃𝑦 𝑥𝑅𝑦} = {𝑥 ∣ ∃𝑦𝜑})
13	1, 12	eqtrid 2783	1 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → dom 𝑅 = {𝑥 ∣ ∃𝑦𝜑})

Syntax hints:   → wi 4   = wceq 1541  ∃wex 1780   ∈ wcel 2113  {cab 2714  ⟨cop 4586   class class class wbr 5098  {copab 5160  dom cdm 5624

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-10 2146  ax-11 2162  ax-12 2184  ax-ext 2708  ax-sep 5241  ax-nul 5251  ax-pr 5377

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2068  df-mo 2539  df-eu 2569  df-clab 2715  df-cleq 2728  df-clel 2811  df-nfc 2885  df-rab 3400  df-v 3442  df-dif 3904  df-un 3906  df-ss 3918  df-nul 4286  df-if 4480  df-sn 4581  df-pr 4583  df-op 4587  df-br 5099  df-opab 5161  df-dm 5634

This theorem is referenced by:  fpwrelmapffslem  32811