Theorem opabrn 32685

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 5843	. 2 ⊢ ran 𝑅 = {𝑦 ∣ ∃𝑥 𝑥𝑅𝑦}
2		nfopab2 5156	. . . 4 ⊢ Ⅎ𝑦{⟨𝑥, 𝑦⟩ ∣ 𝜑}
3	2	nfeq2 2916	. . 3 ⊢ Ⅎ𝑦 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
4		nfopab1 5155	. . . . 5 ⊢ Ⅎ𝑥{⟨𝑥, 𝑦⟩ ∣ 𝜑}
5	4	nfeq2 2916	. . . 4 ⊢ Ⅎ𝑥 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
6		df-br 5086	. . . . 5 ⊢ (𝑥𝑅𝑦 ↔ ⟨𝑥, 𝑦⟩ ∈ 𝑅)
7		eleq2 2825	. . . . . 6 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ ⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑}))
8		opabidw 5479	. . . . . 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 2231	. . 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 1542  ∃wex 1781   ∈ wcel 2114  {cab 2714  ⟨cop 4573   class class class wbr 5085  {copab 5147  ran crn 5632

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2708  ax-sep 5231  ax-pr 5375

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-nf 1786  df-sb 2069  df-mo 2539  df-eu 2569  df-clab 2715  df-cleq 2728  df-clel 2811  df-nfc 2885  df-rab 3390  df-v 3431  df-dif 3892  df-un 3894  df-in 3896  df-ss 3906  df-nul 4274  df-if 4467  df-sn 4568  df-pr 4570  df-op 4574  df-br 5086  df-opab 5148  df-cnv 5639  df-dm 5641  df-rn 5642

This theorem is referenced by:  fpwrelmapffslem  32805