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Theorem n0eldmqseq 38687
Description: The empty set is not an element of a domain quotient. (Contributed by Peter Mazsa, 3-Nov-2018.)
Assertion
Ref Expression
n0eldmqseq ((dom 𝑅 / 𝑅) = 𝐴 → ¬ ∅ ∈ 𝐴)

Proof of Theorem n0eldmqseq
StepHypRef Expression
1 n0eldmqs 38685 . 2 ¬ ∅ ∈ (dom 𝑅 / 𝑅)
2 eleq2 2820 . 2 ((dom 𝑅 / 𝑅) = 𝐴 → (∅ ∈ (dom 𝑅 / 𝑅) ↔ ∅ ∈ 𝐴))
31, 2mtbii 326 1 ((dom 𝑅 / 𝑅) = 𝐴 → ¬ ∅ ∈ 𝐴)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1541  wcel 2111  c0 4278  dom cdm 5611   / cqs 8616
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703  ax-sep 5229  ax-nul 5239  ax-pr 5365
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ne 2929  df-ral 3048  df-rex 3057  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-in 3904  df-ss 3914  df-nul 4279  df-if 4471  df-sn 4572  df-pr 4574  df-op 4578  df-br 5087  df-opab 5149  df-xp 5617  df-cnv 5619  df-dm 5621  df-rn 5622  df-res 5623  df-ima 5624  df-ec 8619  df-qs 8623
This theorem is referenced by:  n0el3  38689  fences3  38868  mainer  38872
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