Theorem fundif 6613

Description: A function with removed elements is still a function. (Contributed by AV, 7-Jun-2021.) Shorten proof and remove dependency on ax-sep 5253 and ax-pr 5396. (Revised by Eric Schmidt, 24-Oct-2026.)

Assertion

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

Proof of Theorem fundif

Step	Hyp	Ref	Expression
1		difss 4075	. 2 ⊢ (𝐹 ∖ 𝐴) ⊆ 𝐹
2		funss 6581	. 2 ⊢ ((𝐹 ∖ 𝐴) ⊆ 𝐹 → (Fun 𝐹 → Fun (𝐹 ∖ 𝐴)))
3	1, 2	ax-mp 5	1 ⊢ (Fun 𝐹 → Fun (𝐹 ∖ 𝐴))

This proof depends on syntax axioms:   → wi 4   ⊆ wss 3900   ∖ cdif 4048  Fun wfun 6555

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-ss 3901  df-dif 4050  df-br 5107  df-opab 5172  df-rel 5663  df-cnv 5664  df-co 5665  df-fun 6563

This theorem is used by:  fundmge2nop  14678  fun2dmnop  14680