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Theorem elrn3 35724
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 5711 . . 3 ran 𝐵 = dom 𝐵
21eleq2i 2836 . 2 (𝐴 ∈ ran 𝐵𝐴 ∈ dom 𝐵)
3 eldm3 35723 . 2 (𝐴 ∈ dom 𝐵 ↔ (𝐵 ↾ {𝐴}) ≠ ∅)
4 cnvxp 6188 . . . . . . 7 (V × {𝐴}) = ({𝐴} × V)
54ineq2i 4238 . . . . . 6 (𝐵(V × {𝐴})) = (𝐵 ∩ ({𝐴} × V))
6 cnvin 6176 . . . . . 6 (𝐵 ∩ (V × {𝐴})) = (𝐵(V × {𝐴}))
7 df-res 5712 . . . . . 6 (𝐵 ↾ {𝐴}) = (𝐵 ∩ ({𝐴} × V))
85, 6, 73eqtr4ri 2779 . . . . 5 (𝐵 ↾ {𝐴}) = (𝐵 ∩ (V × {𝐴}))
98eqeq1i 2745 . . . 4 ((𝐵 ↾ {𝐴}) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
10 relinxp 5838 . . . . 5 Rel (𝐵 ∩ (V × {𝐴}))
11 cnveq0 6228 . . . . 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 2999 . 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 1537  wcel 2108  wne 2946  Vcvv 3488  cin 3975  c0 4352  {csn 4648   × cxp 5698  ccnv 5699  dom cdm 5700  ran crn 5701  cres 5702  Rel wrel 5705
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-11 2158  ax-12 2178  ax-ext 2711  ax-sep 5317  ax-nul 5324  ax-pr 5447
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ne 2947  df-ral 3068  df-rex 3077  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-in 3983  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-br 5167  df-opab 5229  df-xp 5706  df-rel 5707  df-cnv 5708  df-dm 5710  df-rn 5711  df-res 5712
This theorem is referenced by: (None)
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