Metamath Proof Explorer


Theorem sbcbidvOLD

Description: Obsolete version of sbcbidv as of 1-Dec-2023. (Contributed by NM, 29-Dec-2014) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypothesis sbcbidv.1
|- ( ph -> ( ps <-> ch ) )
Assertion sbcbidvOLD
|- ( ph -> ( [. A / x ]. ps <-> [. A / x ]. ch ) )

Proof

Step Hyp Ref Expression
1 sbcbidv.1
 |-  ( ph -> ( ps <-> ch ) )
2 nfv
 |-  F/ x ph
3 2 1 sbcbid
 |-  ( ph -> ( [. A / x ]. ps <-> [. A / x ]. ch ) )