Theorem qseq2 8325

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 8310	. . . . 5 ⊢ (𝐴 = 𝐵 → [𝑥]𝐴 = [𝑥]𝐵)
2	1	eqeq2d 2831	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑦 = [𝑥]𝐴 ↔ 𝑦 = [𝑥]𝐵))
3	2	rexbidv 3292	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴 ↔ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵))
4	3	abbidv 2884	. 2 ⊢ (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴} = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵})
5		df-qs 8276	. 2 ⊢ (𝐶 / 𝐴) = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐴}
6		df-qs 8276	. 2 ⊢ (𝐶 / 𝐵) = {𝑦 ∣ ∃𝑥 ∈ 𝐶 𝑦 = [𝑥]𝐵}
7	4, 5, 6	3eqtr4g 2880	1 ⊢ (𝐴 = 𝐵 → (𝐶 / 𝐴) = (𝐶 / 𝐵))

Syntax hints:   → wi 4   = wceq 1537  {cab 2798  ∃wrex 3134  [cec 8268   / cqs 8269

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 2792

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 2799  df-cleq 2813  df-clel 2891  df-nfc 2959  df-rex 3139  df-rab 3142  df-v 3483  df-dif 3922  df-un 3924  df-in 3926  df-ss 3935  df-nul 4275  df-if 4449  df-sn 4549  df-pr 4551  df-op 4555  df-br 5048  df-opab 5110  df-cnv 5544  df-dm 5546  df-rn 5547  df-res 5548  df-ima 5549  df-ec 8272  df-qs 8276

This theorem is referenced by:  qseq2i  8326  qseq2d  8327  qseq12  8328  pi1bas3  23625  pstmval  31140