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Theorem elrn3 35741
Description: Quantifier-free definition of membership in a range. (Contributed by Scott Fenton, 21-Jan-2017.)
Assertion
Ref Expression
elrn3 (𝐴 ∈ ran 𝐵 ↔ (𝐵 ∩ (V × {𝐴})) ≠ ∅)

Proof of Theorem elrn3
StepHypRef Expression
1 df-rn 5699 . . 3 ran 𝐵 = dom 𝐵
21eleq2i 2830 . 2 (𝐴 ∈ ran 𝐵𝐴 ∈ dom 𝐵)
3 eldm3 35740 . 2 (𝐴 ∈ dom 𝐵 ↔ (𝐵 ↾ {𝐴}) ≠ ∅)
4 cnvxp 6178 . . . . . . 7 (V × {𝐴}) = ({𝐴} × V)
54ineq2i 4224 . . . . . 6 (𝐵(V × {𝐴})) = (𝐵 ∩ ({𝐴} × V))
6 cnvin 6166 . . . . . 6 (𝐵 ∩ (V × {𝐴})) = (𝐵(V × {𝐴}))
7 df-res 5700 . . . . . 6 (𝐵 ↾ {𝐴}) = (𝐵 ∩ ({𝐴} × V))
85, 6, 73eqtr4ri 2773 . . . . 5 (𝐵 ↾ {𝐴}) = (𝐵 ∩ (V × {𝐴}))
98eqeq1i 2739 . . . 4 ((𝐵 ↾ {𝐴}) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
10 relinxp 5826 . . . . 5 Rel (𝐵 ∩ (V × {𝐴}))
11 cnveq0 6218 . . . . 5 (Rel (𝐵 ∩ (V × {𝐴})) → ((𝐵 ∩ (V × {𝐴})) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅))
1210, 11ax-mp 5 . . . 4 ((𝐵 ∩ (V × {𝐴})) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
139, 12bitr4i 278 . . 3 ((𝐵 ↾ {𝐴}) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
1413necon3bii 2990 . 2 ((𝐵 ↾ {𝐴}) ≠ ∅ ↔ (𝐵 ∩ (V × {𝐴})) ≠ ∅)
152, 3, 143bitri 297 1 (𝐴 ∈ ran 𝐵 ↔ (𝐵 ∩ (V × {𝐴})) ≠ ∅)
Colors of variables: wff setvar class
Syntax hints:  wb 206   = wceq 1536  wcel 2105  wne 2937  Vcvv 3477  cin 3961  c0 4338  {csn 4630   × cxp 5686  ccnv 5687  dom cdm 5688  ran crn 5689  cres 5690  Rel wrel 5693
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1791  ax-4 1805  ax-5 1907  ax-6 1964  ax-7 2004  ax-8 2107  ax-9 2115  ax-11 2154  ax-12 2174  ax-ext 2705  ax-sep 5301  ax-nul 5311  ax-pr 5437
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1539  df-fal 1549  df-ex 1776  df-sb 2062  df-clab 2712  df-cleq 2726  df-clel 2813  df-ne 2938  df-ral 3059  df-rex 3068  df-rab 3433  df-v 3479  df-dif 3965  df-un 3967  df-in 3969  df-ss 3979  df-nul 4339  df-if 4531  df-sn 4631  df-pr 4633  df-op 4637  df-br 5148  df-opab 5210  df-xp 5694  df-rel 5695  df-cnv 5696  df-dm 5698  df-rn 5699  df-res 5700
This theorem is referenced by: (None)
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