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	$$⊢Ⅎx\phi$$
	sbbid.2	$$⊢\phi \to \left(\psi \leftrightarrow \chi \right)$$
Assertion	sbbid	$$⊢\phi \to \left(\left[y/x\right]\psi \leftrightarrow \left[y/x\right]\chi \right)$$

Proof

Step	Hyp	Ref	Expression
1		sbbid.1	$$⊢Ⅎx\phi$$
2		sbbid.2	$$⊢\phi \to \left(\psi \leftrightarrow \chi \right)$$
3	1 2	alrimi	$$⊢\phi \to \forall x\left(\psi \leftrightarrow \chi \right)$$
4		spsbbi	$$⊢\forall x\left(\psi \leftrightarrow \chi \right)\to \left(\left[y/x\right]\psi \leftrightarrow \left[y/x\right]\chi \right)$$
5	3 4	syl	$$⊢\phi \to \left(\left[y/x\right]\psi \leftrightarrow \left[y/x\right]\chi \right)$$