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Theorem n0elqs2 34438
 Description: Two ways of expressing that the empty set is not an element of a quotient set. (Contributed by Peter Mazsa, 25-Jul-2021.)
Assertion
Ref Expression
n0elqs2 (¬ ∅ ∈ (𝐴 / 𝑅) ↔ dom (𝑅𝐴) = 𝐴)

Proof of Theorem n0elqs2
StepHypRef Expression
1 n0elqs 34437 . 2 (¬ ∅ ∈ (𝐴 / 𝑅) ↔ 𝐴 ⊆ dom 𝑅)
2 ssdmres 5561 . 2 (𝐴 ⊆ dom 𝑅 ↔ dom (𝑅𝐴) = 𝐴)
31, 2bitri 264 1 (¬ ∅ ∈ (𝐴 / 𝑅) ↔ dom (𝑅𝐴) = 𝐴)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ↔ wb 196   = wceq 1631   ∈ wcel 2145   ⊆ wss 3723  ∅c0 4063  dom cdm 5249   ↾ cres 5251   / cqs 7894 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-9 2154  ax-10 2174  ax-11 2190  ax-12 2203  ax-13 2408  ax-ext 2751  ax-sep 4915  ax-nul 4923  ax-pr 5034 This theorem depends on definitions:  df-bi 197  df-an 383  df-or 827  df-3an 1073  df-tru 1634  df-ex 1853  df-nf 1858  df-sb 2050  df-eu 2622  df-mo 2623  df-clab 2758  df-cleq 2764  df-clel 2767  df-nfc 2902  df-ne 2944  df-ral 3066  df-rex 3067  df-rab 3070  df-v 3353  df-sbc 3588  df-dif 3726  df-un 3728  df-in 3730  df-ss 3737  df-nul 4064  df-if 4226  df-sn 4317  df-pr 4319  df-op 4323  df-br 4787  df-opab 4847  df-xp 5255  df-cnv 5257  df-dm 5259  df-rn 5260  df-res 5261  df-ima 5262  df-ec 7897  df-qs 7901 This theorem is referenced by: (None)
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