Theorem nf5di

Description: Since the converse holds by a1i , this inference shows that we can represent a not-free hypothesis with either F/ x ph (inference form) or ( ph -> F/ x ph ) (deduction form). (Contributed by NM, 17-Aug-2018) (Proof shortened by Wolf Lammen, 10-Jul-2019) Reduce axiom dependencies. (Revised by SN and Sky Winter, 23-Oct-2026)

	Ref	Expression
Hypothesis	nf5di.1	$$⊢\phi \to Ⅎx\phi$$
Assertion	nf5di	$$⊢Ⅎx\phi$$

Proof

Step	Hyp	Ref	Expression
1		nf5di.1	$$⊢\phi \to Ⅎx\phi$$
2		hbe1	$$⊢\exists x\phi \to \forall x\exists x\phi$$
3	1	nfrd	$$⊢\phi \to \left(\exists x\phi \to \forall x\phi \right)$$
4	3	com12	$$⊢\exists x\phi \to \left(\phi \to \forall x\phi \right)$$
5	2 4	eximdh	$$⊢\exists x\phi \to \left(\exists x\phi \to \exists x\forall x\phi \right)$$
6	5	pm2.43i	$$⊢\exists x\phi \to \exists x\forall x\phi$$
7		hbe1a	$$⊢\exists x\forall x\phi \to \forall x\phi$$
8	6 7	syl	$$⊢\exists x\phi \to \forall x\phi$$
9	8	nfi	$$⊢Ⅎx\phi$$