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 Ⅎ' x φ t x φ φ

Proof

Step Hyp Ref Expression
1 spsbe t x φ x φ
2 bj-nnfe Ⅎ' x φ x φ φ
3 1 2 syl5 Ⅎ' x φ t x φ φ
4 bj-nnfa Ⅎ' x φ φ x φ
5 stdpc4 x φ t x φ
6 4 5 syl6 Ⅎ' x φ φ t x φ
7 3 6 impbid Ⅎ' x φ t x φ φ