Metamath Proof Explorer


Theorem sb1

Description: One direction of a simplified definition of substitution. The converse requires either a disjoint variable condition ( sb5 ) or a non-freeness hypothesis ( sb5f ). Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker sb1v when possible. (Contributed by NM, 13-May-1993) Revise df-sb . (Revised by Wolf Lammen, 21-Feb-2024) (New usage is discouraged.)

Ref Expression
Assertion sb1
|- ( [ y / x ] ph -> E. x ( x = y /\ ph ) )

Proof

Step Hyp Ref Expression
1 spsbe
 |-  ( [ y / x ] ph -> E. x ph )
2 pm3.2
 |-  ( x = y -> ( ph -> ( x = y /\ ph ) ) )
3 2 aleximi
 |-  ( A. x x = y -> ( E. x ph -> E. x ( x = y /\ ph ) ) )
4 1 3 syl5
 |-  ( A. x x = y -> ( [ y / x ] ph -> E. x ( x = y /\ ph ) ) )
5 sb3b
 |-  ( -. A. x x = y -> ( [ y / x ] ph <-> E. x ( x = y /\ ph ) ) )
6 5 biimpd
 |-  ( -. A. x x = y -> ( [ y / x ] ph -> E. x ( x = y /\ ph ) ) )
7 4 6 pm2.61i
 |-  ( [ y / x ] ph -> E. x ( x = y /\ ph ) )