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Metamath Proof Explorer

Theorem sb1v

Description: One direction of sb5 , provable from fewer axioms. Version of sb1 with a disjoint variable condition using fewer axioms. (Contributed by NM, 13-May-1993) (Revised by Wolf Lammen, 20-Jan-2024)

		Ref	Expression
	Assertion	sb1v	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		sb6	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )
2		equs4v	⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) → ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) )
3	1 2	sylbi	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) )