Theorem newbdayim

Description: One direction of the biconditional in newbday . (Contributed by Scott Fenton, 7-Nov-2025)

	Ref	Expression
Assertion	newbdayim	$$⊢X\in N\left(A\right)\to \mathrm{bday}\left(X\right)=A$$

Proof

Step	Hyp	Ref	Expression
1		elfvdm	$$⊢X\in N\left(A\right)\to A\in \mathrm{dom}N$$
2		newf	$$⊢N:\mathrm{On}⟶𝒫\mathrm{No}$$
3		fdm	$$⊢N:\mathrm{On}⟶𝒫\mathrm{No}\to \mathrm{dom}N=\mathrm{On}$$
4	2 3	ax-mp	$$⊢\mathrm{dom}N=\mathrm{On}$$
5	1 4	eleqtrdi	$$⊢X\in N\left(A\right)\to A\in \mathrm{On}$$
6		newssno	$$⊢N\left(A\right)\subseteq \mathrm{No}$$
7	6	sseli	$$⊢X\in N\left(A\right)\to X\in \mathrm{No}$$
8		newbday	$$⊢\left(A\in \mathrm{On}\wedge X\in \mathrm{No}\right)\to \left(X\in N\left(A\right)\leftrightarrow \mathrm{bday}\left(X\right)=A\right)$$
9	5 7 8	syl2anc	$$⊢X\in N\left(A\right)\to \left(X\in N\left(A\right)\leftrightarrow \mathrm{bday}\left(X\right)=A\right)$$
10	9	ibi	$$⊢X\in N\left(A\right)\to \mathrm{bday}\left(X\right)=A$$