Theorem qseq2 8778

Description: Equality theorem for quotient set. (Contributed by NM, 23-Jul-1995.)

Assertion

Ref	Expression
qseq2	⊢ (𝐴 = 𝐵 → (𝐶 / 𝐴) = (𝐶 / 𝐵))

Proof of Theorem qseq2

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

Step	Hyp	Ref	Expression
1		eceq2 8759	. . . . 5 ⊢ (𝐴 = 𝐵 → [𝑥]𝐴 = [𝑥]𝐵)
2	1	eqeq2d 2772	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑦 = [𝑥]𝐴 ↔ 𝑦 = [𝑥]𝐵))
3	2	rexbidv 3187	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴 ↔ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵))
4	3	abbidv 2827	. 2 ⊢ (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴} = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵})
5		df-qs 8723	. 2 ⊢ (𝐶 / 𝐴) = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴}
6		df-qs 8723	. 2 ⊢ (𝐶 / 𝐵) = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵}
7	4, 5, 6	3eqtr4g 2821	1 ⊢ (𝐴 = 𝐵 → (𝐶 / 𝐴) = (𝐶 / 𝐵))

This proof depends on syntax axioms:   → wi 4   = wceq 1570  {cab 2739  ∃wrex 3087  [cec 8715   / cqs 8716

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 2733

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 2740  df-cleq 2753  df-clel 2836  df-rex 3088  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-cnv 5659  df-dm 5661  df-rn 5662  df-res 5663  df-ima 5664  df-ec 8719  df-qs 8723

This theorem is used by:  qseq2i  8779  qseq2d  8781  qseq12  8782  pi1bas3  25364