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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	$⊢ [y/ x] φ \to \exists x (x = y \land φ)$

Proof

Step	Hyp	Ref	Expression
1		sb6	$⊢ [y/ x] φ \leftrightarrow \forall x (x = y \to φ)$
2		equs4v	$⊢ \forall x (x = y \to φ) \to \exists x (x = y \land φ)$
3	1 2	sylbi	$⊢ [y/ x] φ \to \exists x (x = y \land φ)$