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 ) C_ F
2 funss
 |-  ( ( F \ A ) C_ F -> ( Fun F -> Fun ( F \ A ) ) )
3 1 2 ax-mp
 |-  ( Fun F -> Fun ( F \ A ) )