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 ↔ ∀ 𝑦 ( 𝑦 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ) )

Proof

Step	Hyp	Ref	Expression
1		sbjust	⊢ ( ∀ 𝑦 ( 𝑦 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ) ↔ ∀ 𝑧 ( 𝑧 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜑 ) ) )
2	1	df-sb	⊢ ( [ 𝑡 / 𝑥 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ) )