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Theorem afvvfveq 41228
Description: The value of the alternative function at a set as argument equals the function's value at this argument. (Contributed by Alexander van der Vekens, 25-May-2017.)
Assertion
Ref Expression
afvvfveq ((𝐹'''𝐴) ∈ 𝐵 → (𝐹'''𝐴) = (𝐹𝐴))

Proof of Theorem afvvfveq
StepHypRef Expression
1 nvelim 41200 . . 3 ((𝐹'''𝐴) = V → ¬ (𝐹'''𝐴) ∈ 𝐵)
21necon2ai 2823 . 2 ((𝐹'''𝐴) ∈ 𝐵 → (𝐹'''𝐴) ≠ V)
3 afvnufveq 41227 . 2 ((𝐹'''𝐴) ≠ V → (𝐹'''𝐴) = (𝐹𝐴))
42, 3syl 17 1 ((𝐹'''𝐴) ∈ 𝐵 → (𝐹'''𝐴) = (𝐹𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1483  wcel 1990  wne 2794  Vcvv 3200  cfv 5888  '''cafv 41194
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-8 1992  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-sep 4781
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-rab 2921  df-v 3202  df-un 3579  df-if 4087  df-fv 5896  df-afv 41197
This theorem is referenced by:  afv0fv0  41229  afv0nbfvbi  41231  aovvoveq  41272
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