Theorem csbeq1 2000

Description: Analog of dfsbcq 1940 for proper substitution into a class.

Assertion

Ref	Expression
csbeq1	|- (A = B -> [_A / x]_C = [_B / x]_C)

Proof of Theorem csbeq1

Step	Hyp	Ref	Expression
1		dfsbcq 1940	. . 3 |- (A = B -> ([A / x]y e. C <-> [B / x]y e. C))
2	1	abbidv 1575	. 2 |- (A = B -> {y | [A / x]y e. C} = {y | [B / x]y e. C})
3		df-csb 1999	. 2 |- [_A / x]_C = {y | [A / x]y e. C}
4		df-csb 1999	. 2 |- [_B / x]_C = {y | [B / x]y e. C}
5	2, 3, 4	3eqtr4g 1529	1 |- (A = B -> [_A / x]_C = [_B / x]_C)

Syntax hints:   -> wi 3   = wceq 955   e. wcel 957  [wsbc 1169  {cab 1462  [_csb 1998

This theorem is referenced by:  csbeq1d 2001  csbeq1a 2003  fsum1slem 6961  csbfsum 6980  fsumshftm 6985  ipval2 8319

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 961  ax-gen 962  ax-8 963  ax-10 965  ax-12 967  ax-17 970  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122  ax-10o 1139  ax-16 1209  ax-11o 1217  ax-ext 1458

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 980  df-sb 1171  df-clab 1463  df-cleq 1468  df-clel 1471  df-sbc 1939  df-csb 1999

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