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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	$⊢ φ \to (ψ \leftrightarrow χ)$
Assertion	sbbid	$⊢ φ \to ([y/ x] ψ \leftrightarrow [y/ x] χ)$

Proof

Step	Hyp	Ref	Expression
1		sbbid.1	$⊢ Ⅎ x φ$
2		sbbid.2	$⊢ φ \to (ψ \leftrightarrow χ)$
3	1 2	alrimi	$⊢ φ \to \forall x (ψ \leftrightarrow χ)$
4		spsbbi	$⊢ \forall x (ψ \leftrightarrow χ) \to ([y/ x] ψ \leftrightarrow [y/ x] χ)$
5	3 4	syl	$⊢ φ \to ([y/ x] ψ \leftrightarrow [y/ x] χ)$