Theorem qseq2 4295

Description: Equality theorem for quotient set.

Assertion

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

Proof of Theorem qseq2

Step	Hyp	Ref	Expression
1		eceq1 4283	. . . . 5 ⊢ (A = B → [x]A = [x]B)
2	1	eqeq2d 1489	. . . 4 ⊢ (A = B → (y = [x]A ↔ y = [x]B))
3	2	rexbidv 1667	. . 3 ⊢ (A = B → (∃x ∈ C y = [x]A ↔ ∃x ∈ C y = [x]B))
4	3	abbidv 1580	. 2 ⊢ (A = B → {y∣∃x ∈ C y = [x]A} = {y∣∃x ∈ C y = [x]B})
5		df-qs 4272	. 2 ⊢ (C / A) = {y∣∃x ∈ C y = [x]A}
6		df-qs 4272	. 2 ⊢ (C / B) = {y∣∃x ∈ C y = [x]B}
7	4, 5, 6	3eqtr4g 1534	1 ⊢ (A = B → (C / A) = (C / B))

Syntax hints:   → wi 3   = wceq 958  {cab 1466  ∃wrex 1649  [cec 4265   / cqs 4266

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-11 969  ax-12 970  ax-13 971  ax-14 972  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462  ax-sep 2708  ax-pow 2748  ax-pr 2785

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174  df-eu 1384  df-mo 1385  df-clab 1467  df-cleq 1472  df-clel 1475  df-ne 1590  df-rex 1653  df-v 1815  df-dif 2052  df-un 2053  df-in 2054  df-ss 2056  df-nul 2284  df-pw 2406  df-sn 2416  df-pr 2417  df-op 2420  df-br 2625  df-opab 2672  df-cnv 3192  df-dm 3194  df-rn 3195  df-res 3196  df-ima 3197  df-ec 4269  df-qs 4272

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