Theorem releldmqs 39782

Description: Elementhood in the domain quotient of a relation. (Contributed by Peter Mazsa, 24-Apr-2021.)

Assertion

Ref	Expression
releldmqs	⊢ (𝐴 ∈ 𝑉 → (Rel 𝑅 → (𝐴 ∈ (dom 𝑅 / 𝑅) ↔ ∃𝑢 ∈ dom 𝑅∃𝑥 ∈ [ 𝑢]𝑅𝐴 = [𝑢]𝑅)))

Distinct variable groups:   𝑢,𝐴,𝑥   𝑢,𝑅,𝑥

Allowed substitution hints:   𝑉(𝑥, 𝑢)

Proof of Theorem releldmqs

Step	Hyp	Ref	Expression
1		resdm 6016	. . . . . 6 ⊢ (Rel 𝑅 → (𝑅 ↾ dom 𝑅) = 𝑅)
2	1	dmqseqd 39765	. . . . 5 ⊢ (Rel 𝑅 → (dom (𝑅 ↾ dom 𝑅) / (𝑅 ↾ dom 𝑅)) = (dom 𝑅 / 𝑅))
3	2	eleq2d 2846	. . . 4 ⊢ (Rel 𝑅 → (𝐴 ∈ (dom (𝑅 ↾ dom 𝑅) / (𝑅 ↾ dom 𝑅)) ↔ 𝐴 ∈ (dom 𝑅 / 𝑅)))
4	3	adantl 487	. . 3 ⊢ ((𝐴 ∈ 𝑉 ∧ Rel 𝑅) → (𝐴 ∈ (dom (𝑅 ↾ dom 𝑅) / (𝑅 ↾ dom 𝑅)) ↔ 𝐴 ∈ (dom 𝑅 / 𝑅)))
5		eldmqsres2 39333	. . . 4 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ (dom (𝑅 ↾ dom 𝑅) / (𝑅 ↾ dom 𝑅)) ↔ ∃𝑢 ∈ dom 𝑅∃𝑥 ∈ [ 𝑢]𝑅𝐴 = [𝑢]𝑅))
6	5	adantr 486	. . 3 ⊢ ((𝐴 ∈ 𝑉 ∧ Rel 𝑅) → (𝐴 ∈ (dom (𝑅 ↾ dom 𝑅) / (𝑅 ↾ dom 𝑅)) ↔ ∃𝑢 ∈ dom 𝑅∃𝑥 ∈ [ 𝑢]𝑅𝐴 = [𝑢]𝑅))
7	4, 6	bitr3d 284	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ Rel 𝑅) → (𝐴 ∈ (dom 𝑅 / 𝑅) ↔ ∃𝑢 ∈ dom 𝑅∃𝑥 ∈ [ 𝑢]𝑅𝐴 = [𝑢]𝑅))
8	7	ex 418	1 ⊢ (𝐴 ∈ 𝑉 → (Rel 𝑅 → (𝐴 ∈ (dom 𝑅 / 𝑅) ↔ ∃𝑢 ∈ dom 𝑅∃𝑥 ∈ [ 𝑢]𝑅𝐴 = [𝑢]𝑅)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2144  ∃wrex 3086  dom cdm 5650   ↾ cres 5652  Rel wrel 5655  [cec 8720   / cqs 8721

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248  ax-pr 5390

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-xp 5656  df-rel 5657  df-cnv 5658  df-dm 5660  df-rn 5661  df-res 5662  df-ima 5663  df-ec 8724  df-qs 8728

This theorem is used by:  disjdmqsss  39944