Theorem erALTVeq1 39793

Description: Equality theorem for equivalence relation on domain quotient. (Contributed by Peter Mazsa, 25-Sep-2021.)

Assertion

Ref	Expression
erALTVeq1	⊢ (𝑅 = 𝑆 → (𝑅 ErALTV 𝐴 ↔ 𝑆 ErALTV 𝐴))

Proof of Theorem erALTVeq1

Step	Hyp	Ref	Expression
1		eqvreleq 39725	. . 3 ⊢ (𝑅 = 𝑆 → ( EqvRel 𝑅 ↔ EqvRel 𝑆))
2		dmqseqeq1 39766	. . 3 ⊢ (𝑅 = 𝑆 → ((dom 𝑅 / 𝑅) = 𝐴 ↔ (dom 𝑆 / 𝑆) = 𝐴))
3	1, 2	anbi12d 644	. 2 ⊢ (𝑅 = 𝑆 → (( EqvRel 𝑅 ∧ (dom 𝑅 / 𝑅) = 𝐴) ↔ ( EqvRel 𝑆 ∧ (dom 𝑆 / 𝑆) = 𝐴)))
4		dferALTV2 39792	. 2 ⊢ (𝑅 ErALTV 𝐴 ↔ ( EqvRel 𝑅 ∧ (dom 𝑅 / 𝑅) = 𝐴))
5		dferALTV2 39792	. 2 ⊢ (𝑆 ErALTV 𝐴 ↔ ( EqvRel 𝑆 ∧ (dom 𝑆 / 𝑆) = 𝐴))
6	3, 4, 5	3bitr4g 317	1 ⊢ (𝑅 = 𝑆 → (𝑅 ErALTV 𝐴 ↔ 𝑆 ErALTV 𝐴))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569  dom cdm 5650   / cqs 8721   EqvRel weqvrel 39239   ErALTV werALTV 39248

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-co 5659  df-dm 5660  df-rn 5661  df-res 5662  df-ima 5663  df-ec 8724  df-qs 8728  df-refrel 39631  df-symrel 39663  df-trrel 39697  df-eqvrel 39708  df-dmqs 39762  df-erALTV 39788

This theorem is used by:  erALTVeq1i  39794  erALTVeq1d  39795