Metamath Proof Explorer


Theorem fnrndomnum

Description: A version of fnrndomg that does not require the axiom of choice ax-ac . (Contributed by Vincent Gonzalez, 17-Aug-2026)

Ref Expression
Assertion fnrndomnum ( 𝐴 ∈ dom card → ( 𝐹 Fn 𝐴 → ran 𝐹𝐴 ) )

Proof

Step Hyp Ref Expression
1 dffn4 ( 𝐹 Fn 𝐴𝐹 : 𝐴onto→ ran 𝐹 )
2 fodomnum ( 𝐴 ∈ dom card → ( 𝐹 : 𝐴onto→ ran 𝐹 → ran 𝐹𝐴 ) )
3 1 2 biimtrid ( 𝐴 ∈ dom card → ( 𝐹 Fn 𝐴 → ran 𝐹𝐴 ) )