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Metamath Proof Explorer

Theorem sbbid

Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993) Remove dependency on ax-10 and ax-13 . (Revised by Wolf Lammen, 24-Nov-2022) Revise df-sb . (Revised by Steven Nguyen, 11-Jul-2023)

	Ref	Expression
Hypotheses	sbbid.1	⊢ Ⅎ 𝑥 𝜑
	sbbid.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion	sbbid	⊢ ( 𝜑 → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ [ 𝑦 / 𝑥 ] 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		sbbid.1	⊢ Ⅎ 𝑥 𝜑
2		sbbid.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
3	1 2	alrimi	⊢ ( 𝜑 → ∀ 𝑥 ( 𝜓 ↔ 𝜒 ) )
4		spsbbi	⊢ ( ∀ 𝑥 ( 𝜓 ↔ 𝜒 ) → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ [ 𝑦 / 𝑥 ] 𝜒 ) )
5	3 4	syl	⊢ ( 𝜑 → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ [ 𝑦 / 𝑥 ] 𝜒 ) )