Metamath Proof Explorer


Theorem bj-subst

Description: Proof of sbalex from core axioms, ax-10 (modal5), and bj-ax12 . (Contributed by BJ, 29-Dec-2020) (Proof modification is discouraged.)

Ref Expression
Assertion bj-subst ( ∃ 𝑥 ( 𝑥 = 𝑦𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )

Proof

Step Hyp Ref Expression
1 bj-ax12 𝑥 ( 𝑥 = 𝑦 → ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ) )
2 pm3.31 ( ( 𝑥 = 𝑦 → ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ) ) → ( ( 𝑥 = 𝑦𝜑 ) → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ) )
3 2 aleximi ( ∀ 𝑥 ( 𝑥 = 𝑦 → ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ) ) → ( ∃ 𝑥 ( 𝑥 = 𝑦𝜑 ) → ∃ 𝑥𝑥 ( 𝑥 = 𝑦𝜑 ) ) )
4 1 3 ax-mp ( ∃ 𝑥 ( 𝑥 = 𝑦𝜑 ) → ∃ 𝑥𝑥 ( 𝑥 = 𝑦𝜑 ) )
5 hbe1a ( ∃ 𝑥𝑥 ( 𝑥 = 𝑦𝜑 ) → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )
6 4 5 syl ( ∃ 𝑥 ( 𝑥 = 𝑦𝜑 ) → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )
7 equs4v ( ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) → ∃ 𝑥 ( 𝑥 = 𝑦𝜑 ) )
8 6 7 impbii ( ∃ 𝑥 ( 𝑥 = 𝑦𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )