Theorem qseq2 8789

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 8770	. . . . 5 ⊢ (𝐴 = 𝐵 → [𝑥]𝐴 = [𝑥]𝐵)
2	1	eqeq2d 2774	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑦 = [𝑥]𝐴 ↔ 𝑦 = [𝑥]𝐵))
3	2	rexbidv 3189	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴 ↔ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵))
4	3	abbidv 2829	. 2 ⊢ (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴} = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵})
5		df-qs 8734	. 2 ⊢ (𝐶 / 𝐴) = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴}
6		df-qs 8734	. 2 ⊢ (𝐶 / 𝐵) = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵}
7	4, 5, 6	3eqtr4g 2823	1 ⊢ (𝐴 = 𝐵 → (𝐶 / 𝐴) = (𝐶 / 𝐵))

This proof depends on syntax axioms:   → wi 4   = wceq 1569  {cab 2741  ∃wrex 3089  [cec 8726   / cqs 8727

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 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

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 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-rex 3090  df-rab 3416  df-v 3455  df-ss 3900  df-dif 4049  df-in 4093  df-un 4153  df-nul 4281  df-if 4485  df-sn 4588  df-pr 4590  df-op 4838  df-br 5106  df-opab 5170  df-cnv 5662  df-dm 5664  df-rn 5665  df-res 5666  df-ima 5667  df-ec 8730  df-qs 8734

This theorem is used by:  qseq2i  8790  qseq2d  8792  qseq12  8793  pi1bas3  25376