Theorem elqsecl 8770

Description: Membership in a quotient set by an equivalence class according to ∼. (Contributed by Alexander van der Vekens, 12-Apr-2018.) (Revised by AV, 30-Apr-2021.)

Assertion

Ref	Expression
elqsecl	⊢ (𝐵 ∈ 𝑋 → (𝐵 ∈ (𝑊 / ∼ ) ↔ ∃𝑥 ∈ 𝑊 𝐵 = {𝑦 ∣ 𝑥 ∼ 𝑦}))

Distinct variable groups:   𝑥, ∼ ,𝑦   𝑥,𝐵   𝑥,𝑊   𝑥,𝑋

Allowed substitution hints:   𝐵(𝑦)   𝑊(𝑦)   𝑋(𝑦)

Proof of Theorem elqsecl

Step	Hyp	Ref	Expression
1		elqsg 8767	. 2 ⊢ (𝐵 ∈ 𝑋 → (𝐵 ∈ (𝑊 / ∼ ) ↔ ∃𝑥 ∈ 𝑊 𝐵 = [𝑥] ∼ ))
2		vex 3457	. . . . 5 ⊢ 𝑥 ∈ V
3		dfec2 8703	. . . . 5 ⊢ (𝑥 ∈ V → [𝑥] ∼ = {𝑦 ∣ 𝑥 ∼ 𝑦})
4	2, 3	mp1i 14	. . . 4 ⊢ (𝐵 ∈ 𝑋 → [𝑥] ∼ = {𝑦 ∣ 𝑥 ∼ 𝑦})
5	4	eqeq2d 2773	. . 3 ⊢ (𝐵 ∈ 𝑋 → (𝐵 = [𝑥] ∼ ↔ 𝐵 = {𝑦 ∣ 𝑥 ∼ 𝑦}))
6	5	rexbidv 3188	. 2 ⊢ (𝐵 ∈ 𝑋 → (∃𝑥 ∈ 𝑊 𝐵 = [𝑥] ∼ ↔ ∃𝑥 ∈ 𝑊 𝐵 = {𝑦 ∣ 𝑥 ∼ 𝑦}))
7	1, 6	bitrd 282	1 ⊢ (𝐵 ∈ 𝑋 → (𝐵 ∈ (𝑊 / ∼ ) ↔ ∃𝑥 ∈ 𝑊 𝐵 = {𝑦 ∣ 𝑥 ∼ 𝑦}))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   ∈ wcel 2145  {cab 2740  ∃wrex 3088  Vcvv 3453   class class class wbr 5107  [cec 8698   / cqs 8699

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734  ax-sep 5255  ax-pr 5402

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-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-br 5108  df-opab 5172  df-xp 5665  df-cnv 5667  df-dm 5669  df-rn 5670  df-res 5671  df-ima 5672  df-ec 8702  df-qs 8706

This theorem is used by:  eclclwwlkn1  30553