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Theorem bj-ssbequ1 31321
Description: This uses ax-12 1950 with a direct reference to ax12v 1951. Therefore, compared to bj-ax12 31311, there is a hidden use of sp 1957. Note that with ax-12 1950, it can be proved with dv condition on  x ,  t. See sbequ1 2097. (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.
StepHypRef Expression
1 equtr2 1877 . . . . . . . 8  |-  ( ( y  =  t  /\  x  =  t )  ->  y  =  x )
21equcomd 1871 . . . . . . 7  |-  ( ( y  =  t  /\  x  =  t )  ->  x  =  y )
3 ax12v 1951 . . . . . . 7  |-  ( x  =  y  ->  ( ph  ->  A. x ( x  =  y  ->  ph )
) )
42, 3syl 17 . . . . . 6  |-  ( ( y  =  t  /\  x  =  t )  ->  ( ph  ->  A. x
( x  =  y  ->  ph ) ) )
54expimpd 614 . . . . 5  |-  ( y  =  t  ->  (
( x  =  t  /\  ph )  ->  A. x ( x  =  y  ->  ph ) ) )
65com12 31 . . . 4  |-  ( ( x  =  t  /\  ph )  ->  ( y  =  t  ->  A. x
( x  =  y  ->  ph ) ) )
76alrimiv 1781 . . 3  |-  ( ( x  =  t  /\  ph )  ->  A. y
( y  =  t  ->  A. x ( x  =  y  ->  ph )
) )
87ex 441 . 2  |-  ( x  =  t  ->  ( ph  ->  A. y ( y  =  t  ->  A. x
( x  =  y  ->  ph ) ) ) )
9 df-ssb 31297 . 2  |-  ([ t/ x]b ph  <->  A. y ( y  =  t  ->  A. x
( x  =  y  ->  ph ) ) )
108, 9syl6ibr 235 1  |-  ( x  =  t  ->  ( ph  -> [ t/ x]b ph ) )
Colors of variables: wff setvar class
Syntax hints:    -> wi 4    /\ wa 376   A.wal 1450  [wssb 31296
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1677  ax-4 1690  ax-5 1766  ax-6 1813  ax-7 1859  ax-12 1950
This theorem depends on definitions:  df-bi 190  df-an 378  df-ex 1672  df-ssb 31297
This theorem is referenced by:  bj-ssbid1  31324
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