Theorem nfbidf

Description: An equality theorem for effectively not free. (Contributed by Mario Carneiro, 4-Oct-2016) df-nf changed. (Revised by Wolf Lammen, 18-Sep-2021)

	Ref	Expression
Hypotheses	albid.1	$$⊢Ⅎx\phi$$
	albid.2	$$⊢\phi \to \left(\psi \leftrightarrow \chi \right)$$
Assertion	nfbidf	$$⊢\phi \to \left(Ⅎx\psi \leftrightarrow Ⅎx\chi \right)$$

Proof

Step	Hyp	Ref	Expression
1		albid.1	$$⊢Ⅎx\phi$$
2		albid.2	$$⊢\phi \to \left(\psi \leftrightarrow \chi \right)$$
3	1 2	exbid	$$⊢\phi \to \left(\exists x\psi \leftrightarrow \exists x\chi \right)$$
4	1 2	albid	$$⊢\phi \to \left(\forall x\psi \leftrightarrow \forall x\chi \right)$$
5	3 4	imbi12d	$$⊢\phi \to \left(\left(\exists x\psi \to \forall x\psi \right)\leftrightarrow \left(\exists x\chi \to \forall x\chi \right)\right)$$
6		df-nf	$$⊢Ⅎx\psi \leftrightarrow \left(\exists x\psi \to \forall x\psi \right)$$
7		df-nf	$$⊢Ⅎx\chi \leftrightarrow \left(\exists x\chi \to \forall x\chi \right)$$
8	5 6 7	3bitr4g	$$⊢\phi \to \left(Ⅎx\psi \leftrightarrow Ⅎx\chi \right)$$