Theorem opabdm 32108

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 5686	. 2 ⊢ dom 𝑅 = {𝑥 ∣ ∃𝑦 𝑥𝑅𝑦}
2		nfopab1 5218	. . . 4 ⊢ Ⅎ𝑥{⟨𝑥, 𝑦⟩ ∣ 𝜑}
3	2	nfeq2 2919	. . 3 ⊢ Ⅎ𝑥 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
4		nfopab2 5219	. . . . 5 ⊢ Ⅎ𝑦{⟨𝑥, 𝑦⟩ ∣ 𝜑}
5	4	nfeq2 2919	. . . 4 ⊢ Ⅎ𝑦 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
6		df-br 5149	. . . . 5 ⊢ (𝑥𝑅𝑦 ↔ ⟨𝑥, 𝑦⟩ ∈ 𝑅)
7		eleq2 2821	. . . . . 6 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (⟨𝑥, 𝑦⟩ ∈ 𝑅 ↔ ⟨𝑥, 𝑦⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑}))
8		opabidw 5524	. . . . . 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 2215	. . 3 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → (∃𝑦 𝑥𝑅𝑦 ↔ ∃𝑦𝜑))
12	3, 11	abbid 2802	. 2 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → {𝑥 ∣ ∃𝑦 𝑥𝑅𝑦} = {𝑥 ∣ ∃𝑦𝜑})
13	1, 12	eqtrid 2783	1 ⊢ (𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝜑} → dom 𝑅 = {𝑥 ∣ ∃𝑦𝜑})

Syntax hints:   → wi 4   = wceq 1540  ∃wex 1780   ∈ wcel 2105  {cab 2708  ⟨cop 4634   class class class wbr 5148  {copab 5210  dom cdm 5676

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2153  ax-12 2170  ax-ext 2702  ax-sep 5299  ax-nul 5306  ax-pr 5427

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1781  df-nf 1785  df-sb 2067  df-mo 2533  df-eu 2562  df-clab 2709  df-cleq 2723  df-clel 2809  df-nfc 2884  df-rab 3432  df-v 3475  df-dif 3951  df-un 3953  df-in 3955  df-ss 3965  df-nul 4323  df-if 4529  df-sn 4629  df-pr 4631  df-op 4635  df-br 5149  df-opab 5211  df-dm 5686

This theorem is referenced by:  fpwrelmapffslem  32225