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Metamath Proof Explorer

Theorem dfsb1

Description: Alternate definition of substitution. Remark 9.1 in Megill p. 447 (p. 15 of the preprint). This was the original definition before df-sb . Note that it does not require dummy variables in its definiens; this is done by having x free in the first conjunct and bound in the second. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by BJ, 9-Jul-2023) Revise df-sb . (Revised by Wolf Lammen, 29-Jul-2023) (New usage is discouraged.)

Ref Expression
Assertion dfsb1 
|- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )

Proof

Step Hyp Ref Expression
1 sbequ2 
 |-  ( x = y -> ( [ y / x ] ph -> ph ) )
2 1 com12 
 |-  ( [ y / x ] ph -> ( x = y -> ph ) )
3 sb1 
 |-  ( [ y / x ] ph -> E. x ( x = y /\ ph ) )
4 2 3 jca 
 |-  ( [ y / x ] ph -> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
5 id 
 |-  ( x = y -> x = y )
6 sbequ1 
 |-  ( x = y -> ( ph -> [ y / x ] ph ) )
7 5 6 embantd 
 |-  ( x = y -> ( ( x = y -> ph ) -> [ y / x ] ph ) )
8 7 sps 
 |-  ( A. x x = y -> ( ( x = y -> ph ) -> [ y / x ] ph ) )
9 8 adantrd 
 |-  ( A. x x = y -> ( ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) -> [ y / x ] ph ) )
10 sb3 
 |-  ( -. A. x x = y -> ( E. x ( x = y /\ ph ) -> [ y / x ] ph ) )
11 10 adantld 
 |-  ( -. A. x x = y -> ( ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) -> [ y / x ] ph ) )
12 9 11 pm2.61i 
 |-  ( ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) -> [ y / x ] ph )
13 4 12 impbii 
 |-  ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )