Theorem ecelqsg 6835

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 2234	. . 3 ⊢ [𝐵]𝑅 = [𝐵]𝑅
2		eceq1 6815	. . . . 5 ⊢ (𝑥 = 𝐵 → [𝑥]𝑅 = [𝐵]𝑅)
3	2	eqeq2d 2246	. . . 4 ⊢ (𝑥 = 𝐵 → ([𝐵]𝑅 = [𝑥]𝑅 ↔ [𝐵]𝑅 = [𝐵]𝑅))
4	3	rspcev 2923	. . 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 6784	. . . 4 ⊢ (𝑅 ∈ 𝑉 → [𝐵]𝑅 ∈ V)
7		elqsg 6832	. . . 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 1398   ∈ wcel 2205  ∃wrex 2523  Vcvv 2815  [cec 6778   / cqs 6779

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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-13 2207  ax-14 2208  ax-ext 2216  ax-sep 4233  ax-pow 4292  ax-pr 4327  ax-un 4559

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-eu 2085  df-mo 2086  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-ral 2527  df-rex 2528  df-v 2817  df-un 3218  df-in 3220  df-ss 3227  df-pw 3676  df-sn 3700  df-pr 3701  df-op 3703  df-uni 3920  df-br 4115  df-opab 4177  df-xp 4760  df-cnv 4762  df-dm 4764  df-rn 4765  df-res 4766  df-ima 4767  df-ec 6782  df-qs 6786

This theorem is referenced by:  ecelqsi  6836  qliftlem  6860  eroprf  6875  quseccl0g  14032