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Theorem afvvfveq 41742
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 41714 . . 3 ((𝐹'''𝐴) = V → ¬ (𝐹'''𝐴) ∈ 𝐵)
21necon2ai 2971 . 2 ((𝐹'''𝐴) ∈ 𝐵 → (𝐹'''𝐴) ≠ V)
3 afvnufveq 41741 . 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 1630  wcel 2144  wne 2942  Vcvv 3349  cfv 6031  '''cafv 41708
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1869  ax-4 1884  ax-5 1990  ax-6 2056  ax-7 2092  ax-8 2146  ax-9 2153  ax-10 2173  ax-11 2189  ax-12 2202  ax-13 2407  ax-ext 2750  ax-sep 4912
This theorem depends on definitions:  df-bi 197  df-an 383  df-or 827  df-tru 1633  df-ex 1852  df-nf 1857  df-sb 2049  df-clab 2757  df-cleq 2763  df-clel 2766  df-nfc 2901  df-ne 2943  df-rab 3069  df-v 3351  df-un 3726  df-if 4224  df-fv 6039  df-afv 41711
This theorem is referenced by:  afv0fv0  41743  afv0nbfvbi  41745  aovvoveq  41786
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