Metamath Proof Explorer


Theorem sbequ

Description: Equality property for substitution, from Tarski's system. Used in proof of Theorem 9.7 in Megill p. 449 (p. 16 of the preprint). (Contributed by NM, 14-May-1993) Revise df-sb . (Revised by BJ, 30-Dec-2020)

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

Proof

Step Hyp Ref Expression
1 equequ2
 |-  ( x = y -> ( u = x <-> u = y ) )
2 1 imbi1d
 |-  ( x = y -> ( ( u = x -> A. z ( z = u -> ph ) ) <-> ( u = y -> A. z ( z = u -> ph ) ) ) )
3 2 albidv
 |-  ( x = y -> ( A. u ( u = x -> A. z ( z = u -> ph ) ) <-> A. u ( u = y -> A. z ( z = u -> ph ) ) ) )
4 df-sb
 |-  ( [ x / z ] ph <-> A. u ( u = x -> A. z ( z = u -> ph ) ) )
5 df-sb
 |-  ( [ y / z ] ph <-> A. u ( u = y -> A. z ( z = u -> ph ) ) )
6 3 4 5 3bitr4g
 |-  ( x = y -> ( [ x / z ] ph <-> [ y / z ] ph ) )