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Theorem afvvfveq 39877
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 39849 . . 3 ((𝐹'''𝐴) = V → ¬ (𝐹'''𝐴) ∈ 𝐵)
21necon2ai 2811 . 2 ((𝐹'''𝐴) ∈ 𝐵 → (𝐹'''𝐴) ≠ V)
3 afvnufveq 39876 . 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 1475  wcel 1977  wne 2780  Vcvv 3173  cfv 5804  '''cafv 39843
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-8 1979  ax-9 1986  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590  ax-sep 4709
This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-rab 2905  df-v 3175  df-un 3545  df-if 4037  df-fv 5812  df-afv 39846
This theorem is referenced by:  afv0fv0  39878  afv0nbfvbi  39880  aovvoveq  39921
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