Metamath Proof Explorer

Theorem frege55lem1c

Description: Necessary deduction regarding substitution of value in equality. (Contributed by RP, 24-Dec-2019)

Ref Expression
Assertion frege55lem1c 
|- ( ( ph -> [. A / x ]. x = B ) -> ( ph -> A = B ) )

Proof

Step Hyp Ref Expression
1 df-sbc 
 |-  ( [. A / x ]. x = B <-> A e. { x | x = B } )
2 eqeq1 
 |-  ( x = A -> ( x = B <-> A = B ) )
3 2 elabg 
 |-  ( A e. { x | x = B } -> ( A e. { x | x = B } <-> A = B ) )
4 3 ibi 
 |-  ( A e. { x | x = B } -> A = B )
5 1 4 sylbi 
 |-  ( [. A / x ]. x = B -> A = B )
6 5 imim2i 
 |-  ( ( ph -> [. A / x ]. x = B ) -> ( ph -> A = B ) )