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Metamath Proof Explorer

Theorem dfsb

Description: Simplify definition df-sb by removing its provable hypothesis. (Contributed by Wolf Lammen, 5-Feb-2026)

		Ref	Expression
	Assertion	dfsb	$⊢ [t/ x] φ \leftrightarrow \forall y (y = t \to \forall x (x = y \to φ))$

Step	Hyp	Ref	Expression
1		sbjust	$⊢ \forall y (y = t \to \forall x (x = y \to φ)) \leftrightarrow \forall z (z = t \to \forall x (x = z \to φ))$
2	1	df-sb	$⊢ [t/ x] φ \leftrightarrow \forall y (y = t \to \forall x (x = y \to φ))$