Theorem opabrn 30049

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 5652	. 2 ⊢ ran 𝑅 = {𝑦 ∣ ∃𝑥 𝑥𝑅𝑦}
2		nfopab2 5038	. . . 4 ⊢ Ⅎ𝑦{⟨𝑥, 𝑦⟩ ∣ 𝜑}
3	2	nfeq2 2966	. . 3 ⊢ Ⅎ𝑦 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
4		nfopab1 5037	. . . . 5 ⊢ Ⅎ𝑥{⟨𝑥, 𝑦⟩ ∣ 𝜑}
5	4	nfeq2 2966	. . . 4 ⊢ Ⅎ𝑥 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
6		df-br 4969	. . . . 5 ⊢ (𝑥𝑅𝑦 ↔ ⟨𝑥, 𝑦⟩ ∈ 𝑅)
7		eleq2 2873	. . . . . 6 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ ⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑}))
8		opabid 5310	. . . . . 6 ⊢ (⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ↔ 𝜑)
9	7, 8	syl6bb 288	. . . . 5 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ 𝜑))
10	6, 9	syl5bb 284	. . . 4 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (𝑥𝑅𝑦 ↔ 𝜑))
11	5, 10	exbid 2192	. . 3 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (∃𝑥 𝑥𝑅𝑦 ↔ ∃𝑥𝜑))
12	3, 11	abbid 2864	. 2 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → {𝑦 ∣ ∃𝑥 𝑥𝑅𝑦} = {𝑦 ∣ ∃𝑥𝜑})
13	1, 12	syl5eq 2845	1 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → ran 𝑅 = {𝑦 ∣ ∃𝑥𝜑})

Syntax hints:   → wi 4   = wceq 1525  ∃wex 1765   ∈ wcel 2083  {cab 2777  ⟨cop 4484   class class class wbr 4968  {copab 5030  ran crn 5451

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1781  ax-4 1795  ax-5 1892  ax-6 1951  ax-7 1996  ax-8 2085  ax-9 2093  ax-10 2114  ax-11 2128  ax-12 2143  ax-13 2346  ax-ext 2771  ax-sep 5101  ax-nul 5108  ax-pr 5228

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-3an 1082  df-tru 1528  df-ex 1766  df-nf 1770  df-sb 2045  df-mo 2578  df-eu 2614  df-clab 2778  df-cleq 2790  df-clel 2865  df-nfc 2937  df-rab 3116  df-v 3442  df-dif 3868  df-un 3870  df-in 3872  df-ss 3880  df-nul 4218  df-if 4388  df-sn 4479  df-pr 4481  df-op 4485  df-br 4969  df-opab 5031  df-cnv 5458  df-dm 5460  df-rn 5461

This theorem is referenced by:  fpwrelmapffslem  30152