Theorem qseq2 4289

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 4277	. . . . 5 |- (A = B -> [x]A = [x]B)
2	1	eqeq2d 1486	. . . 4 |- (A = B -> (y = [x]A <-> y = [x]B))
3	2	rexbidv 1664	. . 3 |- (A = B -> (E.x e. C y = [x]A <-> E.x e. C y = [x]B))
4	3	abbidv 1577	. 2 |- (A = B -> {y | E.x e. C y = [x]A} = {y | E.x e. C y = [x]B})
5		df-qs 4266	. 2 |- (C/.A) = {y | E.x e. C y = [x]A}
6		df-qs 4266	. 2 |- (C/.B) = {y | E.x e. C y = [x]B}
7	4, 5, 6	3eqtr4g 1531	1 |- (A = B -> (C/.A) = (C/.B))

Syntax hints:   -> wi 3   = wceq 956  {cab 1463  E.wrex 1646  [cec 4259  /.cqs 4260

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-11 967  ax-12 968  ax-13 969  ax-14 970  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459  ax-sep 2703  ax-pow 2742  ax-pr 2779

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172  df-eu 1382  df-mo 1383  df-clab 1464  df-cleq 1469  df-clel 1472  df-ne 1587  df-rex 1650  df-v 1812  df-dif 2049  df-un 2050  df-in 2051  df-ss 2053  df-nul 2281  df-pw 2402  df-sn 2412  df-pr 2413  df-op 2416  df-br 2620  df-opab 2667  df-cnv 3186  df-dm 3188  df-rn 3189  df-res 3190  df-ima 3191  df-ec 4263  df-qs 4266

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