Theorem opabrn 33260

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 5874	. 2 ⊢ ran 𝑅 = {𝑦 ∣ ∃𝑥 𝑥𝑅𝑦}
2		nfopab2 5178	. . . 4 ⊢ Ⅎ𝑦{⟨𝑥, 𝑦⟩ ∣ 𝜑}
3	2	nfeq2 2942	. . 3 ⊢ Ⅎ𝑦 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
4		nfopab1 5177	. . . . 5 ⊢ Ⅎ𝑥{⟨𝑥, 𝑦⟩ ∣ 𝜑}
5	4	nfeq2 2942	. . . 4 ⊢ Ⅎ𝑥 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
6		df-br 5106	. . . . 5 ⊢ (𝑥𝑅𝑦 ↔ ⟨𝑥, 𝑦⟩ ∈ 𝑅)
7		eleq2 2852	. . . . . 6 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ ⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑}))
8		opabidw 5501	. . . . . 6 ⊢ (⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ↔ 𝜑)
9	7, 8	bitrdi 290	. . . . 5 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ 𝜑))
10	6, 9	bitrid 286	. . . 4 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (𝑥𝑅𝑦 ↔ 𝜑))
11	5, 10	exbid 2262	. . 3 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (∃𝑥 𝑥𝑅𝑦 ↔ ∃𝑥𝜑))
12	3, 11	abbid 2831	. 2 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → {𝑦 ∣ ∃𝑥 𝑥𝑅𝑦} = {𝑦 ∣ ∃𝑥𝜑})
13	1, 12	eqtrid 2810	1 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → ran 𝑅 = {𝑦 ∣ ∃𝑥𝜑})

This proof depends on syntax axioms:   → wi 4   = wceq 1569  ∃wex 1811   ∈ wcel 2146  {cab 2741  ⟨cop 4837   class class class wbr 5105  {copab 5169  ran crn 5655

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2196  ax-12 2215  ax-ext 2735  ax-sep 5251  ax-pr 5394

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-nfc 2912  df-rab 3416  df-v 3455  df-ss 3900  df-dif 4049  df-in 4093  df-un 4153  df-nul 4281  df-if 4485  df-sn 4588  df-pr 4590  df-op 4838  df-br 5106  df-opab 5170  df-cnv 5662  df-dm 5664  df-rn 5665

This theorem is used by:  fpwrelmapffslem  33377