Theorem opabdm 33303

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

Assertion

Ref	Expression
opabdm	⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → dom 𝑅 = {𝑥 ∣ ∃𝑦𝜑})

Distinct variable group:   𝑥,𝑦,𝑅

Allowed substitution hints:   𝜑(𝑥, 𝑦)

Proof of Theorem opabdm

Step	Hyp	Ref	Expression
1		df-dm 5668	. 2 ⊢ dom 𝑅 = {𝑥 ∣ ∃𝑦 𝑥𝑅𝑦}
2		nfopab1 5181	. . . 4 ⊢ Ⅎ𝑥{⟨𝑥, 𝑦⟩ ∣ 𝜑}
3	2	nfeq2 2943	. . 3 ⊢ Ⅎ𝑥 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
4		nfopab2 5182	. . . . 5 ⊢ Ⅎ𝑦{⟨𝑥, 𝑦⟩ ∣ 𝜑}
5	4	nfeq2 2943	. . . 4 ⊢ Ⅎ𝑦 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
6		df-br 5109	. . . . 5 ⊢ (𝑥𝑅𝑦 ↔ ⟨𝑥, 𝑦⟩ ∈ 𝑅)
7		eleq2 2853	. . . . . 6 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ ⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑}))
8		opabidw 5505	. . . . . 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 2263	. . 3 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (∃𝑦 𝑥𝑅𝑦 ↔ ∃𝑦𝜑))
12	3, 11	abbid 2831	. 2 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → {𝑥 ∣ ∃𝑦 𝑥𝑅𝑦} = {𝑥 ∣ ∃𝑦𝜑})
13	1, 12	eqtrid 2810	1 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → dom 𝑅 = {𝑥 ∣ ∃𝑦𝜑})

This proof depends on syntax axioms:   → wi 4   = wceq 1570  ∃wex 1812   ∈ wcel 2147  {cab 2741  ⟨cop 4840   class class class wbr 5108  {copab 5173  dom cdm 5658

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5255  ax-pr 5398

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-br 5109  df-opab 5174  df-dm 5668

This theorem is used by:  fpwrelmapffslem  33420