Metamath Proof Explorer


Theorem fundif

Description: A function with removed elements is still a function. (Contributed by AV, 7-Jun-2021) Shorten proof and remove dependency on ax-sep and ax-pr . (Revised by Eric Schmidt, 24-Oct-2026)

Ref Expression
Assertion fundif ( Fun 𝐹 → Fun ( 𝐹 ∖ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 difss ⊢ ( 𝐹 ∖ 𝐴 ) ⊆ 𝐹
2 funss ⊢ ( ( 𝐹 ∖ 𝐴 ) ⊆ 𝐹 → ( Fun 𝐹 → Fun ( 𝐹 ∖ 𝐴 ) ) )
3 1 2 ax-mp ⊢ ( Fun 𝐹 → Fun ( 𝐹 ∖ 𝐴 ) )