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