Metamath Proof Explorer


Theorem newbdayim

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

Ref Expression
Assertion newbdayim ( 𝑋 ∈ ( N ‘ 𝐴 ) → ( bday 𝑋 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 elfvdm ( 𝑋 ∈ ( N ‘ 𝐴 ) → 𝐴 ∈ dom N )
2 newf N : On ⟶ 𝒫 No
3 fdm ( N : On ⟶ 𝒫 No → dom N = On )
4 2 3 ax-mp dom N = On
5 1 4 eleqtrdi ( 𝑋 ∈ ( N ‘ 𝐴 ) → 𝐴 ∈ On )
6 newssno ( N ‘ 𝐴 ) ⊆ No
7 6 sseli ( 𝑋 ∈ ( N ‘ 𝐴 ) → 𝑋 No )
8 newbday ( ( 𝐴 ∈ On ∧ 𝑋 No ) → ( 𝑋 ∈ ( N ‘ 𝐴 ) ↔ ( bday 𝑋 ) = 𝐴 ) )
9 5 7 8 syl2anc ( 𝑋 ∈ ( N ‘ 𝐴 ) → ( 𝑋 ∈ ( N ‘ 𝐴 ) ↔ ( bday 𝑋 ) = 𝐴 ) )
10 9 ibi ( 𝑋 ∈ ( N ‘ 𝐴 ) → ( bday 𝑋 ) = 𝐴 )