Theorem bj-ssbequ1 32644

Description: This uses ax-12 2047 with a direct reference to ax12v 2048. Therefore, compared to bj-ax12 32634, there is a hidden use of sp 2053. Note that with ax-12 2047, it can be proved with dv condition on x , t. See sbequ1 2110. (Contributed by BJ, 22-Dec-2020.)

Assertion

Ref	Expression
bj-ssbequ1	|- ( x = t -> ( ph -> [ t/ x]b ph ) )

Proof of Theorem bj-ssbequ1

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equtr2 1954	. . . . . . . 8 |- ( ( y = t /\ x = t ) -> y = x )
2	1	equcomd 1946	. . . . . . 7 |- ( ( y = t /\ x = t ) -> x = y )
3		ax12v 2048	. . . . . . 7 |- ( x = y -> ( ph -> A. x ( x = y -> ph ) ) )
4	2, 3	syl 17	. . . . . 6 |- ( ( y = t /\ x = t ) -> ( ph -> A. x ( x = y -> ph ) ) )
5	4	expimpd 629	. . . . 5 |- ( y = t -> ( ( x = t /\ ph ) -> A. x ( x = y -> ph ) ) )
6	5	com12 32	. . . 4 |- ( ( x = t /\ ph ) -> ( y = t -> A. x ( x = y -> ph ) ) )
7	6	alrimiv 1855	. . 3 |- ( ( x = t /\ ph ) -> A. y ( y = t -> A. x ( x = y -> ph ) ) )
8	7	ex 450	. 2 |- ( x = t -> ( ph -> A. y ( y = t -> A. x ( x = y -> ph ) ) ) )
9		df-ssb 32620	. 2 |- ([ t/ x]b ph <-> A. y ( y = t -> A. x ( x = y -> ph ) ) )
10	8, 9	syl6ibr 242	1 |- ( x = t -> ( ph -> [ t/ x]b ph ) )

Syntax hints:   -> wi 4   /\ wa 384   A.wal 1481  [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  ax-7 1935  ax-12 2047

This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705  df-ssb 32620

This theorem is referenced by:  bj-ssbid1  32647