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Theorem elrn3 36254
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 5672 . . 3 ran 𝐵 = dom 𝐵
21eleq2i 2855 . 2 (𝐴 ∈ ran 𝐵𝐴 ∈ dom 𝐵)
3 eldm3 36253 . 2 (𝐴 ∈ dom 𝐵 ↔ (𝐵 ↾ {𝐴}) ≠ ∅)
4 cnvxp 6154 . . . . . . 7 (V × {𝐴}) = ({𝐴} × V)
54ineq2i 4170 . . . . . 6 (𝐵(V × {𝐴})) = (𝐵 ∩ ({𝐴} × V))
6 cnvin 6141 . . . . . 6 (𝐵 ∩ (V × {𝐴})) = (𝐵(V × {𝐴}))
7 df-res 5673 . . . . . 6 (𝐵 ↾ {𝐴}) = (𝐵 ∩ ({𝐴} × V))
85, 6, 73eqtr4ri 2797 . . . . 5 (𝐵 ↾ {𝐴}) = (𝐵 ∩ (V × {𝐴}))
98eqeq1i 2768 . . . 4 ((𝐵 ↾ {𝐴}) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
10 relinxp 5801 . . . . 5 Rel (𝐵 ∩ (V × {𝐴}))
11 cnveq0 6196 . . . . 5 (Rel (𝐵 ∩ (V × {𝐴})) → ((𝐵 ∩ (V × {𝐴})) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅))
1210, 11ax-mp 5 . . . 4 ((𝐵 ∩ (V × {𝐴})) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
139, 12bitr4i 281 . . 3 ((𝐵 ↾ {𝐴}) = ∅ ↔ (𝐵 ∩ (V × {𝐴})) = ∅)
1413necon3bii 3010 . 2 ((𝐵 ↾ {𝐴}) ≠ ∅ ↔ (𝐵 ∩ (V × {𝐴})) ≠ ∅)
152, 3, 143bitri 300 1 (𝐴 ∈ ran 𝐵 ↔ (𝐵 ∩ (V × {𝐴})) ≠ ∅)
Colors of variables: wff setvar class
Syntax hints:  wb 209   = wceq 1570  wcel 2143  wne 2958  Vcvv 3455  cin 3904  c0 4286  {csn 4589   × cxp 5659  ccnv 5660  dom cdm 5661  ran crn 5662  cres 5663  Rel wrel 5666
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-11 2192  ax-ext 2735  ax-sep 5257  ax-pr 5404
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ne 2959  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4488  df-sn 4590  df-pr 4592  df-op 4596  df-br 5110  df-opab 5174  df-xp 5667  df-rel 5668  df-cnv 5669  df-dm 5671  df-rn 5672  df-res 5673
This theorem is referenced by: (None)
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