Theorem opabrn 32597

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

Assertion

Ref	Expression
opabrn	⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → ran 𝑅 = {𝑦 ∣ ∃𝑥𝜑})

Distinct variable group:   𝑥,𝑦,𝑅

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem opabrn

Step	Hyp	Ref	Expression
1		dfrn2 5873	. 2 ⊢ ran 𝑅 = {𝑦 ∣ ∃𝑥 𝑥𝑅𝑦}
2		nfopab2 5195	. . . 4 ⊢ Ⅎ𝑦{⟨𝑥, 𝑦⟩ ∣ 𝜑}
3	2	nfeq2 2917	. . 3 ⊢ Ⅎ𝑦 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
4		nfopab1 5194	. . . . 5 ⊢ Ⅎ𝑥{⟨𝑥, 𝑦⟩ ∣ 𝜑}
5	4	nfeq2 2917	. . . 4 ⊢ Ⅎ𝑥 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
6		df-br 5125	. . . . 5 ⊢ (𝑥𝑅𝑦 ↔ ⟨𝑥, 𝑦⟩ ∈ 𝑅)
7		eleq2 2824	. . . . . 6 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ ⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑}))
8		opabidw 5504	. . . . . 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 2224	. . 3 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (∃𝑥 𝑥𝑅𝑦 ↔ ∃𝑥𝜑))
12	3, 11	abbid 2804	. 2 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → {𝑦 ∣ ∃𝑥 𝑥𝑅𝑦} = {𝑦 ∣ ∃𝑥𝜑})
13	1, 12	eqtrid 2783	1 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → ran 𝑅 = {𝑦 ∣ ∃𝑥𝜑})

Syntax hints:   → wi 4   = wceq 1540  ∃wex 1779   ∈ wcel 2109  {cab 2714  ⟨cop 4612   class class class wbr 5124  {copab 5186  ran crn 5660

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2708  ax-sep 5271  ax-nul 5281  ax-pr 5407

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2540  df-eu 2569  df-clab 2715  df-cleq 2728  df-clel 2810  df-nfc 2886  df-rab 3421  df-v 3466  df-dif 3934  df-un 3936  df-ss 3948  df-nul 4314  df-if 4506  df-sn 4607  df-pr 4609  df-op 4613  df-br 5125  df-opab 5187  df-cnv 5667  df-dm 5669  df-rn 5670

This theorem is referenced by:  fpwrelmapffslem  32714