Theorem bj-sbeqALT 32895

Description: Substitution in an equality (use the more genereal version bj-sbeq 32896 instead, without disjoint variable condition). (Contributed by BJ, 6-Oct-2018.) (New usage is discouraged.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-sbeqALT	|- ( [ y / x ] A = B <-> [_ y / x ]_ A = [_ y / x ]_ B )

Distinct variable group:   x, y

Allowed substitution hints:   A( x, y)   B( x, y)

Proof of Theorem bj-sbeqALT

Step	Hyp	Ref	Expression
1		nfcsb1v 3549	. . 3 |- F/_ x [_ y / x ]_ A
2		nfcsb1v 3549	. . 3 |- F/_ x [_ y / x ]_ B
3	1, 2	nfeq 2776	. 2 |- F/ x [_ y / x ]_ A = [_ y / x ]_ B
4		csbeq1a 3542	. . 3 |- ( x = y -> A = [_ y / x ]_ A )
5		csbeq1a 3542	. . 3 |- ( x = y -> B = [_ y / x ]_ B )
6	4, 5	eqeq12d 2637	. 2 |- ( x = y -> ( A = B <-> [_ y / x ]_ A = [_ y / x ]_ B ) )
7	3, 6	sbie 2408	1 |- ( [ y / x ] A = B <-> [_ y / x ]_ A = [_ y / x ]_ B )

Syntax hints:   <-> wb 196   = wceq 1483   [wsb 1880   [_csb 3533

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-sbc 3436  df-csb 3534

This theorem is referenced by: (None)