Theorem ecelqsg 6862

Description: Membership of an equivalence class in a quotient set. (Contributed by Jeff Madsen, 10-Jun-2010.) (Revised by Mario Carneiro, 9-Jul-2014.)

Assertion

Ref	Expression
ecelqsg	⊢ ((𝑅 ∈ 𝑉 ∧ 𝐵 ∈ 𝐴) → [𝐵]𝑅 ∈ (𝐴 / 𝑅))

Proof of Theorem ecelqsg

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eqid 2238	. . 3 ⊢ [𝐵]𝑅 = [𝐵]𝑅
2		eceq1 6842	. . . . 5 ⊢ (𝑥 = 𝐵 → [𝑥]𝑅 = [𝐵]𝑅)
3	2	eqeq2d 2250	. . . 4 ⊢ (𝑥 = 𝐵 → ([𝐵]𝑅 = [𝑥]𝑅 ↔ [𝐵]𝑅 = [𝐵]𝑅))
4	3	rspcev 2929	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ [𝐵]𝑅 = [𝐵]𝑅) → ∃𝑥 ∈ 𝐴 [𝐵]𝑅 = [𝑥]𝑅)
5	1, 4	mpan2 429	. 2 ⊢ (𝐵 ∈ 𝐴 → ∃𝑥 ∈ 𝐴 [𝐵]𝑅 = [𝑥]𝑅)
6		ecexg 6811	. . . 4 ⊢ (𝑅 ∈ 𝑉 → [𝐵]𝑅 ∈ V)
7		elqsg 6859	. . . 4 ⊢ ([𝐵]𝑅 ∈ V → ([𝐵]𝑅 ∈ (𝐴 / 𝑅) ↔ ∃𝑥 ∈ 𝐴 [𝐵]𝑅 = [𝑥]𝑅))
8	6, 7	syl 14	. . 3 ⊢ (𝑅 ∈ 𝑉 → ([𝐵]𝑅 ∈ (𝐴 / 𝑅) ↔ ∃𝑥 ∈ 𝐴 [𝐵]𝑅 = [𝑥]𝑅))
9	8	biimpar 297	. 2 ⊢ ((𝑅 ∈ 𝑉 ∧ ∃𝑥 ∈ 𝐴 [𝐵]𝑅 = [𝑥]𝑅) → [𝐵]𝑅 ∈ (𝐴 / 𝑅))
10	5, 9	sylan2 286	1 ⊢ ((𝑅 ∈ 𝑉 ∧ 𝐵 ∈ 𝐴) → [𝐵]𝑅 ∈ (𝐴 / 𝑅))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1402   ∈ wcel 2209  ∃wrex 2529  Vcvv 2821  [cec 6805   / cqs 6806

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4249  ax-pow 4311  ax-pr 4346  ax-un 4578

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-un 3224  df-in 3226  df-ss 3233  df-pw 3690  df-sn 3715  df-pr 3716  df-op 3718  df-uni 3936  df-br 4131  df-opab 4193  df-xp 4780  df-cnv 4782  df-dm 4784  df-rn 4785  df-res 4786  df-ima 4787  df-ec 6809  df-qs 6813

This theorem is used by:  ecelqsi  6863  qliftlem  6887  eroprf  6902  quseccl0g  14036