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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 = 𝑦 ∧ 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )

Proof

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