Metamath Proof Explorer


Theorem bj-sbft

Description: Version of sbft using F// , proved from core axioms. (Contributed by BJ, 19-Nov-2023)

Ref Expression
Assertion bj-sbft
|- ( F// x ph -> ( [ t / x ] ph <-> ph ) )

Proof

Step Hyp Ref Expression
1 spsbe
 |-  ( [ t / x ] ph -> E. x ph )
2 bj-nnfe
 |-  ( F// x ph -> ( E. x ph -> ph ) )
3 1 2 syl5
 |-  ( F// x ph -> ( [ t / x ] ph -> ph ) )
4 bj-nnfa
 |-  ( F// x ph -> ( ph -> A. x ph ) )
5 stdpc4
 |-  ( A. x ph -> [ t / x ] ph )
6 4 5 syl6
 |-  ( F// x ph -> ( ph -> [ t / x ] ph ) )
7 3 6 impbid
 |-  ( F// x ph -> ( [ t / x ] ph <-> ph ) )