Theorem ecelqsg 6647

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 2196	. . 3 ⊢ [𝐵]𝑅 = [𝐵]𝑅
2		eceq1 6627	. . . . 5 ⊢ (𝑥 = 𝐵 → [𝑥]𝑅 = [𝐵]𝑅)
3	2	eqeq2d 2208	. . . 4 ⊢ (𝑥 = 𝐵 → ([𝐵]𝑅 = [𝑥]𝑅 ↔ [𝐵]𝑅 = [𝐵]𝑅))
4	3	rspcev 2868	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ [𝐵]𝑅 = [𝐵]𝑅) → ∃𝑥 ∈ 𝐴 [𝐵]𝑅 = [𝑥]𝑅)
5	1, 4	mpan2 425	. 2 ⊢ (𝐵 ∈ 𝐴 → ∃𝑥 ∈ 𝐴 [𝐵]𝑅 = [𝑥]𝑅)
6		ecexg 6596	. . . 4 ⊢ (𝑅 ∈ 𝑉 → [𝐵]𝑅 ∈ V)
7		elqsg 6644	. . . 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))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1364   ∈ wcel 2167  ∃wrex 2476  Vcvv 2763  [cec 6590   / cqs 6591

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-13 2169  ax-14 2170  ax-ext 2178  ax-sep 4151  ax-pow 4207  ax-pr 4242  ax-un 4468

This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-nf 1475  df-sb 1777  df-eu 2048  df-mo 2049  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-ral 2480  df-rex 2481  df-v 2765  df-un 3161  df-in 3163  df-ss 3170  df-pw 3607  df-sn 3628  df-pr 3629  df-op 3631  df-uni 3840  df-br 4034  df-opab 4095  df-xp 4669  df-cnv 4671  df-dm 4673  df-rn 4674  df-res 4675  df-ima 4676  df-ec 6594  df-qs 6598

This theorem is referenced by:  ecelqsi  6648  qliftlem  6672  eroprf  6687  quseccl0g  13361