Theorem bj-ssbn 32641

Description: The result of a substitution in the negation of a formula is the negation of the result of the same substitution in that formula. Proved from Tarski, ax-10 2019, bj-ax12 32634. Compare sbn 2391. (Contributed by BJ, 25-Dec-2020.)

Assertion

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

Proof of Theorem bj-ssbn

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		alinexa 1770	. . . 4 |- ( A. x ( x = y -> -. ph ) <-> -. E. x ( x = y /\ ph ) )
3	2	imbi2i 326	. . 3 |- ( ( y = t -> A. x ( x = y -> -. ph ) ) <-> ( y = t -> -. E. x ( x = y /\ ph ) ) )
4	3	albii 1747	. 2 |- ( A. y ( y = t -> A. x ( x = y -> -. ph ) ) <-> A. y ( y = t -> -. E. x ( x = y /\ ph ) ) )
5		alinexa 1770	. . 3 |- ( A. y ( y = t -> -. E. x ( x = y /\ ph ) ) <-> -. E. y ( y = t /\ E. x ( x = y /\ ph ) ) )
6		bj-dfssb2 32640	. . 3 |- ([ t/ x]b ph <-> E. y ( y = t /\ E. x ( x = y /\ ph ) ) )
7	5, 6	xchbinxr 325	. 2 |- ( A. y ( y = t -> -. E. x ( x = y /\ ph ) ) <-> -. [ t/ x]b ph )
8	1, 4, 7	3bitri 286	1 |- ([ t/ x]b -. ph <-> -. [ t/ x]b ph )

Syntax hints:   -. wn 3   -> wi 4   <-> wb 196   /\ wa 384   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  ax-7 1935  ax-10 2019  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: (None)