Theorem bj-ssb0 32628

Description: Substitution for a variable not occurring in a proposition. See sbf 2380. (Contributed by BJ, 22-Dec-2020.)

Assertion

Ref	Expression
bj-ssb0	|- ([ t/ x]b ph <-> ph )

Distinct variable group:   ph, x

Allowed substitution hint:   ph( t)

Proof of Theorem bj-ssb0

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-ssb 32620	. 2 |- ([ t/ x]b ph <-> A. y ( y = t -> A. x ( x = y -> ph ) ) )
2		19.23v 1902	. . . . . . 7 |- ( A. x ( x = y -> ph ) <-> ( E. x x = y -> ph ) )
3		ax6ev 1890	. . . . . . . . 9 |- E. x x = y
4		pm2.27 42	. . . . . . . . 9 |- ( E. x x = y -> ( ( E. x x = y -> ph ) -> ph ) )
5	3, 4	ax-mp 5	. . . . . . . 8 |- ( ( E. x x = y -> ph ) -> ph )
6		ax-1 6	. . . . . . . 8 |- ( ph -> ( E. x x = y -> ph ) )
7	5, 6	impbii 199	. . . . . . 7 |- ( ( E. x x = y -> ph ) <-> ph )
8	2, 7	bitri 264	. . . . . 6 |- ( A. x ( x = y -> ph ) <-> ph )
9	8	imbi2i 326	. . . . 5 |- ( ( y = t -> A. x ( x = y -> ph ) ) <-> ( y = t -> ph ) )
10	9	albii 1747	. . . 4 |- ( A. y ( y = t -> A. x ( x = y -> ph ) ) <-> A. y ( y = t -> ph ) )
11		19.23v 1902	. . . 4 |- ( A. y ( y = t -> ph ) <-> ( E. y y = t -> ph ) )
12	10, 11	bitri 264	. . 3 |- ( A. y ( y = t -> A. x ( x = y -> ph ) ) <-> ( E. y y = t -> ph ) )
13		ax6ev 1890	. . . . 5 |- E. y y = t
14		pm2.27 42	. . . . 5 |- ( E. y y = t -> ( ( E. y y = t -> ph ) -> ph ) )
15	13, 14	ax-mp 5	. . . 4 |- ( ( E. y y = t -> ph ) -> ph )
16		ax-1 6	. . . 4 |- ( ph -> ( E. y y = t -> ph ) )
17	15, 16	impbii 199	. . 3 |- ( ( E. y y = t -> ph ) <-> ph )
18	12, 17	bitri 264	. 2 |- ( A. y ( y = t -> A. x ( x = y -> ph ) ) <-> ph )
19	1, 18	bitri 264	1 |- ([ t/ x]b ph <-> ph )

Syntax hints:   -> wi 4   <-> wb 196   A.wal 1481   E.wex 1704  [wssb 32619

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

This theorem depends on definitions:  df-bi 197  df-ex 1705  df-ssb 32620

This theorem is referenced by: (None)