Theorem csbeq1 2003

Description: Analog of dfsbcq 1943 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 1943	. . 3 |- (A = B -> ([A / x]y e. C <-> [B / x]y e. C))
2	1	abbidv 1577	. 2 |- (A = B -> {y | [A / x]y e. C} = {y | [B / x]y e. C})
3		df-csb 2002	. 2 |- [_A / x]_C = {y | [A / x]y e. C}
4		df-csb 2002	. 2 |- [_B / x]_C = {y | [B / x]y e. C}
5	2, 3, 4	3eqtr4g 1531	1 |- (A = B -> [_A / x]_C = [_B / x]_C)

Syntax hints:   -> wi 3   = wceq 956   e. wcel 958  [wsbc 1170  {cab 1463  [_csb 2001

This theorem is referenced by:  csbeq1d 2004  csbeq1a 2006  fsum1slem 7008  csbfsum 7027  fsumshftm 7032  ipval2 8357

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-12 968  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

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-sbc 1942  df-csb 2002

Copyright terms: Public domain