Theorem brdmqss 39582

Description: The domain quotient binary relation. (Contributed by Peter Mazsa, 17-Apr-2019.)

Assertion

Ref	Expression
brdmqss	⊢ ((𝐴 ∈ 𝑉 ∧ 𝑅 ∈ 𝑊) → (𝑅 DomainQss 𝐴 ↔ (dom 𝑅 / 𝑅) = 𝐴))

Proof of Theorem brdmqss

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		dmqseq 39576	. . . 4 ⊢ (𝑥 = 𝑅 → (dom 𝑥 / 𝑥) = (dom 𝑅 / 𝑅))
2		id 23	. . . 4 ⊢ (𝑦 = 𝐴 → 𝑦 = 𝐴)
3	1, 2	eqeqan12d 2774	. . 3 ⊢ ((𝑥 = 𝑅 ∧ 𝑦 = 𝐴) → ((dom 𝑥 / 𝑥) = 𝑦 ↔ (dom 𝑅 / 𝑅) = 𝐴))
4		df-dmqss 39574	. . 3 ⊢ DomainQss = {⟨𝑥, 𝑦⟩ ∣ (dom 𝑥 / 𝑥) = 𝑦}
5	3, 4	brabga 5504	. 2 ⊢ ((𝑅 ∈ 𝑊 ∧ 𝐴 ∈ 𝑉) → (𝑅 DomainQss 𝐴 ↔ (dom 𝑅 / 𝑅) = 𝐴))
6	5	ancoms 464	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝑅 ∈ 𝑊) → (𝑅 DomainQss 𝐴 ↔ (dom 𝑅 / 𝑅) = 𝐴))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145   class class class wbr 5102  dom cdm 5647   / cqs 8694   DomainQss cdmqss 39058

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  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 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-br 5103  df-opab 5167  df-cnv 5655  df-dm 5657  df-rn 5658  df-res 5659  df-ima 5660  df-ec 8697  df-qs 8701  df-dmqss 39574

This theorem is used by:  brdmqssqs  39583  cnvepresdmqss  39589  brparts2  39727  dfpeters2  39826