Theorem 0nelqs 3234

Description: A quotient set doesn't contain the empty set.

Hypotheses

Ref	Expression
0nelqs.1	⊢ Er R
0nelqs.2	⊢ dom R = A

Assertion

Ref	Expression
0nelqs	⊢ ¬ ∅ ∈ (A / R)

Proof of Theorem 0nelqs

Step	Hyp	Ref	Expression
1		visset 1350	. . . . . . 7 ⊢ x ∈ V
2	1	ecdmn0 3217	. . . . . 6 ⊢ (x ∈ dom R ↔ ¬ [x]R = ∅)
3		0nelqs.2	. . . . . . 7 ⊢ dom R = A
4	3	eleq2i 1153	. . . . . 6 ⊢ (x ∈ dom R ↔ x ∈ A)
5		cleqcom 1103	. . . . . . 7 ⊢ ([x]R = ∅ ↔ ∅ = [x]R)
6	5	negbii 162	. . . . . 6 ⊢ (¬ [x]R = ∅ ↔ ¬ ∅ = [x]R)
7	2, 4, 6	3bitr3 156	. . . . 5 ⊢ (x ∈ A ↔ ¬ ∅ = [x]R)
8	7	biimp 133	. . . 4 ⊢ (x ∈ A → ¬ ∅ = [x]R)
9		imnan 207	. . . 4 ⊢ ((x ∈ A → ¬ ∅ = [x]R) ↔ ¬ (x ∈ A ∧ ∅ = [x]R))
10	8, 9	mpbi 164	. . 3 ⊢ ¬ (x ∈ A ∧ ∅ = [x]R)
11	10	nex 779	. 2 ⊢ ¬ ∃x(x ∈ A ∧ ∅ = [x]R)
12		elqsi 3228	. 2 ⊢ (∅ ∈ (A / R) → ∃x(x ∈ A ∧ ∅ = [x]R))
13	11, 12	mto 93	1 ⊢ ¬ ∅ ∈ (A / R)

Syntax hints:  ¬ wn 1   → wi 2   ∧ wa 196  ∃wex 678   = wceq 1091   ∈ wcel 1092  ∅c0 1707  dom cdm 2410  Er wer 3197  [cec 3198   / cqs 3199

This theorem is referenced by:  ecelqsdm 3235  0npq 3844  0nsr 3982

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-opab 2098  df-xp 2424  df-cnv 2426  df-dm 2428  df-rn 2429  df-res 2430  df-ima 2431  df-ec 3202  df-qs 3205

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