Theorem opabrn 32624

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 5899	. 2 ⊢ ran 𝑅 = {𝑦 ∣ ∃𝑥 𝑥𝑅𝑦}
2		nfopab2 5214	. . . 4 ⊢ Ⅎ𝑦{⟨𝑥, 𝑦⟩ ∣ 𝜑}
3	2	nfeq2 2923	. . 3 ⊢ Ⅎ𝑦 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
4		nfopab1 5213	. . . . 5 ⊢ Ⅎ𝑥{⟨𝑥, 𝑦⟩ ∣ 𝜑}
5	4	nfeq2 2923	. . . 4 ⊢ Ⅎ𝑥 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
6		df-br 5144	. . . . 5 ⊢ (𝑥𝑅𝑦 ↔ ⟨𝑥, 𝑦⟩ ∈ 𝑅)
7		eleq2 2830	. . . . . 6 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ ⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑}))
8		opabidw 5529	. . . . . 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 2223	. . 3 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (∃𝑥 𝑥𝑅𝑦 ↔ ∃𝑥𝜑))
12	3, 11	abbid 2810	. 2 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → {𝑦 ∣ ∃𝑥 𝑥𝑅𝑦} = {𝑦 ∣ ∃𝑥𝜑})
13	1, 12	eqtrid 2789	1 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → ran 𝑅 = {𝑦 ∣ ∃𝑥𝜑})

Syntax hints:   → wi 4   = wceq 1540  ∃wex 1779   ∈ wcel 2108  {cab 2714  ⟨cop 4632   class class class wbr 5143  {copab 5205  ran crn 5686

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 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708  ax-sep 5296  ax-nul 5306  ax-pr 5432

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-mo 2540  df-eu 2569  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-br 5144  df-opab 5206  df-cnv 5693  df-dm 5695  df-rn 5696

This theorem is referenced by:  fpwrelmapffslem  32743