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Theorem elrn3 35584
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 5693 . . 3 ran 𝐵 = dom 𝐵
21eleq2i 2818 . 2 (𝐴 ∈ ran 𝐵𝐴 ∈ dom 𝐵)
3 eldm3 35583 . 2 (𝐴 ∈ dom 𝐵 ↔ (𝐵 ↾ {𝐴}) ≠ ∅)
4 cnvxp 6168 . . . . . . 7 (V × {𝐴}) = ({𝐴} × V)
54ineq2i 4210 . . . . . 6 (𝐵(V × {𝐴})) = (𝐵 ∩ ({𝐴} × V))
6 cnvin 6156 . . . . . 6 (𝐵 ∩ (V × {𝐴})) = (𝐵(V × {𝐴}))
7 df-res 5694 . . . . . 6 (𝐵 ↾ {𝐴}) = (𝐵 ∩ ({𝐴} × V))
85, 6, 73eqtr4ri 2765 . . . . 5 (𝐵 ↾ {𝐴}) = (𝐵 ∩ (V × {𝐴}))
98eqeq1i 2731 . . . 4 ((𝐵 ↾ {𝐴}) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
10 relinxp 5820 . . . . 5 Rel (𝐵 ∩ (V × {𝐴}))
11 cnveq0 6208 . . . . 5 (Rel (𝐵 ∩ (V × {𝐴})) → ((𝐵 ∩ (V × {𝐴})) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅))
1210, 11ax-mp 5 . . . 4 ((𝐵 ∩ (V × {𝐴})) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
139, 12bitr4i 277 . . 3 ((𝐵 ↾ {𝐴}) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
1413necon3bii 2983 . 2 ((𝐵 ↾ {𝐴}) ≠ ∅ ↔ (𝐵 ∩ (V × {𝐴})) ≠ ∅)
152, 3, 143bitri 296 1 (𝐴 ∈ ran 𝐵 ↔ (𝐵 ∩ (V × {𝐴})) ≠ ∅)
Colors of variables: wff setvar class
Syntax hints:  wb 205   = wceq 1534  wcel 2099  wne 2930  Vcvv 3462  cin 3946  c0 4325  {csn 4633   × cxp 5680  ccnv 5681  dom cdm 5682  ran crn 5683  cres 5684  Rel wrel 5687
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-11 2147  ax-12 2167  ax-ext 2697  ax-sep 5304  ax-nul 5311  ax-pr 5433
This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-3an 1086  df-tru 1537  df-fal 1547  df-ex 1775  df-sb 2061  df-clab 2704  df-cleq 2718  df-clel 2803  df-ne 2931  df-ral 3052  df-rex 3061  df-rab 3420  df-v 3464  df-dif 3950  df-un 3952  df-in 3954  df-ss 3964  df-nul 4326  df-if 4534  df-sn 4634  df-pr 4636  df-op 4640  df-br 5154  df-opab 5216  df-xp 5688  df-rel 5689  df-cnv 5690  df-dm 5692  df-rn 5693  df-res 5694
This theorem is referenced by: (None)
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