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 newno ⊢ ( 𝑋 ∈ ( N ‘ 𝐴 ) → 𝑋 ∈ No )
7 newbday ⊢ ( ( 𝐴 ∈ On ∧ 𝑋 ∈ No ) → ( 𝑋 ∈ ( N ‘ 𝐴 ) ↔ ( bday ‘ 𝑋 ) = 𝐴 ) )
8 5 6 7 syl2anc ⊢ ( 𝑋 ∈ ( N ‘ 𝐴 ) → ( 𝑋 ∈ ( N ‘ 𝐴 ) ↔ ( bday ‘ 𝑋 ) = 𝐴 ) )
9 8 ibi ⊢ ( 𝑋 ∈ ( N ‘ 𝐴 ) → ( bday ‘ 𝑋 ) = 𝐴 )