Theorem qseq2 8344

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 8329	. . . . 5 ⊢ (𝐴 = 𝐵 → [𝑥]𝐴 = [𝑥]𝐵)
2	1	eqeq2d 2832	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑦 = [𝑥]𝐴 ↔ 𝑦 = [𝑥]𝐵))
3	2	rexbidv 3297	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴 ↔ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵))
4	3	abbidv 2885	. 2 ⊢ (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴} = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵})
5		df-qs 8295	. 2 ⊢ (𝐶 / 𝐴) = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴}
6		df-qs 8295	. 2 ⊢ (𝐶 / 𝐵) = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵}
7	4, 5, 6	3eqtr4g 2881	1 ⊢ (𝐴 = 𝐵 → (𝐶 / 𝐴) = (𝐶 / 𝐵))

Syntax hints:   → wi 4   = wceq 1537  {cab 2799  ∃wrex 3139  [cec 8287   / cqs 8288

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-rex 3144  df-rab 3147  df-v 3496  df-dif 3939  df-un 3941  df-in 3943  df-ss 3952  df-nul 4292  df-if 4468  df-sn 4568  df-pr 4570  df-op 4574  df-br 5067  df-opab 5129  df-cnv 5563  df-dm 5565  df-rn 5566  df-res 5567  df-ima 5568  df-ec 8291  df-qs 8295

This theorem is referenced by:  qseq2i  8345  qseq2d  8346  qseq12  8347  pi1bas3  23647  pstmval  31135