Metamath Proof Explorer


Theorem bj-sbievw

Description: Lemma for substitution. Closed form of equsalvw and sbievw . (Contributed by BJ, 23-Jul-2023)

Ref Expression
Assertion bj-sbievw ( [ 𝑦 / 𝑥 ] ( 𝜑𝜓 ) → ( [ 𝑦 / 𝑥 ] 𝜑𝜓 ) )

Proof

Step Hyp Ref Expression
1 sb6 ( [ 𝑦 / 𝑥 ] ( 𝜑𝜓 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → ( 𝜑𝜓 ) ) )
2 bj-sblem ( ∀ 𝑥 ( 𝑥 = 𝑦 → ( 𝜑𝜓 ) ) → ( ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ↔ ( ∃ 𝑥 𝑥 = 𝑦𝜓 ) ) )
3 sb6 ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )
4 ax6ev 𝑥 𝑥 = 𝑦
5 4 a1bi ( 𝜓 ↔ ( ∃ 𝑥 𝑥 = 𝑦𝜓 ) )
6 2 3 5 3bitr4g ( ∀ 𝑥 ( 𝑥 = 𝑦 → ( 𝜑𝜓 ) ) → ( [ 𝑦 / 𝑥 ] 𝜑𝜓 ) )
7 1 6 sylbi ( [ 𝑦 / 𝑥 ] ( 𝜑𝜓 ) → ( [ 𝑦 / 𝑥 ] 𝜑𝜓 ) )