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Theorem n0elqs 38583
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 8698 . . 3 (𝑥 ∈ dom 𝑅 ↔ [𝑥]𝑅 ≠ ∅)
21ralbii 3084 . 2 (∀𝑥𝐴 𝑥 ∈ dom 𝑅 ↔ ∀𝑥𝐴 [𝑥]𝑅 ≠ ∅)
3 dfss3 3924 . 2 (𝐴 ⊆ dom 𝑅 ↔ ∀𝑥𝐴 𝑥 ∈ dom 𝑅)
4 nne 2937 . . . . 5 (¬ [𝑥]𝑅 ≠ ∅ ↔ [𝑥]𝑅 = ∅)
54rexbii 3085 . . . 4 (∃𝑥𝐴 ¬ [𝑥]𝑅 ≠ ∅ ↔ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
65notbii 320 . . 3 (¬ ∃𝑥𝐴 ¬ [𝑥]𝑅 ≠ ∅ ↔ ¬ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
7 dfral2 3089 . . 3 (∀𝑥𝐴 [𝑥]𝑅 ≠ ∅ ↔ ¬ ∃𝑥𝐴 ¬ [𝑥]𝑅 ≠ ∅)
8 0ex 5254 . . . . . 6 ∅ ∈ V
98elqs 8713 . . . . 5 (∅ ∈ (𝐴 / 𝑅) ↔ ∃𝑥𝐴 ∅ = [𝑥]𝑅)
10 eqcom 2744 . . . . . 6 (∅ = [𝑥]𝑅 ↔ [𝑥]𝑅 = ∅)
1110rexbii 3085 . . . . 5 (∃𝑥𝐴 ∅ = [𝑥]𝑅 ↔ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
129, 11bitri 275 . . . 4 (∅ ∈ (𝐴 / 𝑅) ↔ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
1312notbii 320 . . 3 (¬ ∅ ∈ (𝐴 / 𝑅) ↔ ¬ ∃𝑥𝐴 [𝑥]𝑅 = ∅)
146, 7, 133bitr4ri 304 . 2 (¬ ∅ ∈ (𝐴 / 𝑅) ↔ ∀𝑥𝐴 [𝑥]𝑅 ≠ ∅)
152, 3, 143bitr4ri 304 1 (¬ ∅ ∈ (𝐴 / 𝑅) ↔ 𝐴 ⊆ dom 𝑅)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 206   = wceq 1542  wcel 2114  wne 2933  wral 3052  wrex 3062  wss 3903  c0 4287  dom cdm 5632  [cec 8643   / cqs 8644
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5243  ax-nul 5253  ax-pr 5379
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-ral 3053  df-rex 3063  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101  df-opab 5163  df-xp 5638  df-cnv 5640  df-dm 5642  df-rn 5643  df-res 5644  df-ima 5645  df-ec 8647  df-qs 8651
This theorem is referenced by:  n0elqs2  38584  n0eldmqs  38983
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