Metamath Proof Explorer

Theorem frege55lem1b

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

Ref Expression
Assertion frege55lem1b 
|- ( ( ph -> [ x / y ] y = z ) -> ( ph -> x = z ) )

Proof

Step Hyp Ref Expression
1 equsb3 
 |-  ( [ x / y ] y = z <-> x = z )
2 1 biimpi 
 |-  ( [ x / y ] y = z -> x = z )
3 2 imim2i 
 |-  ( ( ph -> [ x / y ] y = z ) -> ( ph -> x = z ) )