Theorem sbimd

Description: Deduction substituting both sides of an implication. (Contributed by Wolf Lammen, 24-Nov-2022) Revise df-sb . (Revised by Steven Nguyen, 9-Jul-2023)

	Ref	Expression
Hypotheses	sbimd.1	$$⊢Ⅎx\phi$$
	sbimd.2	$$⊢\phi \to \left(\psi \to \chi \right)$$
Assertion	sbimd	$$⊢\phi \to \left(\left[y/x\right]\psi \to \left[y/x\right]\chi \right)$$

Proof

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