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Theorem fvconstr2 48894
Description: Two ways of expressing 𝐴𝑅𝐵. (Contributed by Zhi Wang, 18-Sep-2024.)
Hypotheses
Ref Expression
fvconstr.1 (𝜑𝐹 = (𝑅 × {𝑌}))
fvconstr2.2 (𝜑𝑋 ∈ (𝐴𝐹𝐵))
Assertion
Ref Expression
fvconstr2 (𝜑𝐴𝑅𝐵)

Proof of Theorem fvconstr2
StepHypRef Expression
1 fvconstr2.2 . . . 4 (𝜑𝑋 ∈ (𝐴𝐹𝐵))
21ne0d 4292 . . 3 (𝜑 → (𝐴𝐹𝐵) ≠ ∅)
3 fvconstr.1 . . . . . . 7 (𝜑𝐹 = (𝑅 × {𝑌}))
43oveqd 7363 . . . . . 6 (𝜑 → (𝐴𝐹𝐵) = (𝐴(𝑅 × {𝑌})𝐵))
5 df-ov 7349 . . . . . 6 (𝐴(𝑅 × {𝑌})𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩)
64, 5eqtrdi 2782 . . . . 5 (𝜑 → (𝐴𝐹𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩))
76neeq1d 2987 . . . 4 (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ ↔ ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅))
8 dmxpss 6118 . . . . 5 dom (𝑅 × {𝑌}) ⊆ 𝑅
9 ndmfv 6854 . . . . . 6 (¬ ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}) → ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) = ∅)
109necon1ai 2955 . . . . 5 (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}))
118, 10sselid 3932 . . . 4 (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
127, 11biimtrdi 253 . . 3 (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅))
132, 12mpd 15 . 2 (𝜑 → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
14 df-br 5092 . 2 (𝐴𝑅𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑅)
1513, 14sylibr 234 1 (𝜑𝐴𝑅𝐵)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1541  wcel 2111  wne 2928  c0 4283  {csn 4576  cop 4582   class class class wbr 5091   × cxp 5614  dom cdm 5616  cfv 6481  (class class class)co 7346
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-11 2160  ax-12 2180  ax-ext 2703  ax-sep 5234  ax-nul 5244  ax-pr 5370
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-mo 2535  df-eu 2564  df-clab 2710  df-cleq 2723  df-clel 2806  df-ne 2929  df-ral 3048  df-rex 3057  df-rab 3396  df-v 3438  df-dif 3905  df-un 3907  df-ss 3919  df-nul 4284  df-if 4476  df-sn 4577  df-pr 4579  df-op 4583  df-uni 4860  df-br 5092  df-opab 5154  df-xp 5622  df-rel 5623  df-cnv 5624  df-dm 5626  df-iota 6437  df-fv 6489  df-ov 7349
This theorem is referenced by:  prsthinc  49495
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