Theorem sbbidv

Description: Deduction substituting both sides of a biconditional, with ph and x disjoint. See also sbbid . (Contributed by Wolf Lammen, 6-May-2023) (Proof shortened by Steven Nguyen, 6-Jul-2023)

	Ref	Expression
Hypothesis	sbbidv.1	$$⊢\phi \to \left(\psi \leftrightarrow \chi \right)$$
Assertion	sbbidv	$$⊢\phi \to \left(\left[t/x\right]\psi \leftrightarrow \left[t/x\right]\chi \right)$$

Proof

Step	Hyp	Ref	Expression
1		sbbidv.1	$$⊢\phi \to \left(\psi \leftrightarrow \chi \right)$$
2	1	alrimiv	$$⊢\phi \to \forall x\left(\psi \leftrightarrow \chi \right)$$
3		spsbbi	$$⊢\forall x\left(\psi \leftrightarrow \chi \right)\to \left(\left[t/x\right]\psi \leftrightarrow \left[t/x\right]\chi \right)$$
4	2 3	syl	$$⊢\phi \to \left(\left[t/x\right]\psi \leftrightarrow \left[t/x\right]\chi \right)$$