Metamath Proof Explorer


Theorem sb2

Description: One direction of a simplified definition of substitution. The converse requires either a disjoint variable condition ( sb6 ) or a non-freeness hypothesis ( sb6f ). Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 13-May-1993) Revise df-sb . (Revised by Wolf Lammen, 26-Jul-2023) (New usage is discouraged.)

Ref Expression
Assertion sb2
|- ( A. x ( x = y -> ph ) -> [ y / x ] ph )

Proof

Step Hyp Ref Expression
1 pm2.27
 |-  ( x = y -> ( ( x = y -> ph ) -> ph ) )
2 1 al2imi
 |-  ( A. x x = y -> ( A. x ( x = y -> ph ) -> A. x ph ) )
3 stdpc4
 |-  ( A. x ph -> [ y / x ] ph )
4 2 3 syl6
 |-  ( A. x x = y -> ( A. x ( x = y -> ph ) -> [ y / x ] ph ) )
5 sb4b
 |-  ( -. A. x x = y -> ( [ y / x ] ph <-> A. x ( x = y -> ph ) ) )
6 5 biimprd
 |-  ( -. A. x x = y -> ( A. x ( x = y -> ph ) -> [ y / x ] ph ) )
7 4 6 pm2.61i
 |-  ( A. x ( x = y -> ph ) -> [ y / x ] ph )