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Theorem n0elqs 38828
Description: Two ways of expressing that the empty set is not an element of a quotient set. (Contributed by Peter Mazsa, 5-Dec-2019.)
Assertion
Ref Expression
n0elqs (¬ ∅ ∈ (𝐴 / 𝑅) ↔ 𝐴 ⊆ dom 𝑅)

Proof of Theorem n0elqs
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 ecdmn0 8731 . . 3 (𝑥 ∈ dom 𝑅 ↔ [𝑥]𝑅 ≠ ∅)
21ralbii 3108 . 2 (∀𝑥𝐴 𝑥 ∈ dom 𝑅 ↔ ∀𝑥𝐴 [𝑥]𝑅 ≠ ∅)
3 dfss3 3925 . 2 (𝐴 ⊆ dom 𝑅 ↔ ∀𝑥𝐴 𝑥 ∈ dom 𝑅)
4 nne 2961 . . . . 5 (¬ [𝑥]𝑅 ≠ ∅ ↔ [𝑥]𝑅 = ∅)
54rexbii 3109 . . . 4 (∃𝑥𝐴 ¬ [𝑥]𝑅 ≠ ∅ ↔ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
65notbii 322 . . 3 (¬ ∃𝑥𝐴 ¬ [𝑥]𝑅 ≠ ∅ ↔ ¬ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
7 dfral2 3113 . . 3 (∀𝑥𝐴 [𝑥]𝑅 ≠ ∅ ↔ ¬ ∃𝑥𝐴 ¬ [𝑥]𝑅 ≠ ∅)
8 0ex 5257 . . . . . 6 ∅ ∈ V
98elqs 8746 . . . . 5 (∅ ∈ (𝐴 / 𝑅) ↔ ∃𝑥𝐴 ∅ = [𝑥]𝑅)
10 eqcom 2769 . . . . . 6 (∅ = [𝑥]𝑅 ↔ [𝑥]𝑅 = ∅)
1110rexbii 3109 . . . . 5 (∃𝑥𝐴 ∅ = [𝑥]𝑅 ↔ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
129, 11bitri 277 . . . 4 (∅ ∈ (𝐴 / 𝑅) ↔ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
1312notbii 322 . . 3 (¬ ∅ ∈ (𝐴 / 𝑅) ↔ ¬ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
146, 7, 133bitr4ri 306 . 2 (¬ ∅ ∈ (𝐴 / 𝑅) ↔ ∀𝑥𝐴 [𝑥]𝑅 ≠ ∅)
152, 3, 143bitr4ri 306 1 (¬ ∅ ∈ (𝐴 / 𝑅) ↔ 𝐴 ⊆ dom 𝑅)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 208   = wceq 1560  wcel 2142  wne 2957  wral 3076  wrex 3086  wss 3904  c0 4285  dom cdm 5647  [cec 8676   / cqs 8677
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734  ax-sep 5246  ax-nul 5256  ax-pr 5390
This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1100  df-tru 1563  df-fal 1573  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-ral 3077  df-rex 3087  df-rab 3415  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4481  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101  df-opab 5163  df-xp 5653  df-cnv 5655  df-dm 5657  df-rn 5658  df-res 5659  df-ima 5660  df-ec 8680  df-qs 8684
This theorem is referenced by:  n0elqs2  38829  n0eldmqs  39228
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