Theorem qseq2 3226

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 3214	. . . . 5 |- (A = B -> [x]A = [x]B)
2	1	cleq2d 1112	. . . 4 |- (A = B -> (y = [x]A <-> y = [x]B))
3	2	birexdv 1220	. . 3 |- (A = B -> (E.x e. C y = [x]A <-> E.x e. C y = [x]B))
4	3	biabdv 1183	. 2 |- (A = B -> {y | E.x e. C y = [x]A} = {y | E.x e. C y = [x]B})
5		df-qs 3205	. 2 |- (C/.A) = {y | E.x e. C y = [x]A}
6		df-qs 3205	. 2 |- (C/.B) = {y | E.x e. C y = [x]B}
7	4, 5, 6	3eqtr4g 1147	1 |- (A = B -> (C/.A) = (C/.B))

Syntax hints:   -> wi 2  {cab 1090   = wceq 1091  E.wrex 1202  [cec 3198  /.cqs 3199

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-opab 2098  df-cnv 2426  df-dm 2428  df-rn 2429  df-res 2430  df-ima 2431  df-ec 3202  df-qs 3205

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