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Theorem fvconstr2 49351
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 4283 . . 3 (𝜑 → (𝐴𝐹𝐵) ≠ ∅)
3 fvconstr.1 . . . . . . 7 (𝜑𝐹 = (𝑅 × {𝑌}))
43oveqd 7377 . . . . . 6 (𝜑 → (𝐴𝐹𝐵) = (𝐴(𝑅 × {𝑌})𝐵))
5 df-ov 7363 . . . . . 6 (𝐴(𝑅 × {𝑌})𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩)
64, 5eqtrdi 2788 . . . . 5 (𝜑 → (𝐴𝐹𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩))
76neeq1d 2992 . . . 4 (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ ↔ ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅))
8 dmxpss 6129 . . . . 5 dom (𝑅 × {𝑌}) ⊆ 𝑅
9 ndmfv 6866 . . . . . 6 (¬ ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}) → ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) = ∅)
109necon1ai 2960 . . . . 5 (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}))
118, 10sselid 3920 . . . 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 5087 . 2 (𝐴𝑅𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑅)
1513, 14sylibr 234 1 (𝜑𝐴𝑅𝐵)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1542  wcel 2114  wne 2933  c0 4274  {csn 4568  cop 4574   class class class wbr 5086   × cxp 5622  dom cdm 5624  cfv 6492  (class class class)co 7360
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5231  ax-nul 5241  ax-pr 5370
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-mo 2540  df-eu 2570  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-ral 3053  df-rex 3063  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-uni 4852  df-br 5087  df-opab 5149  df-xp 5630  df-dm 5634  df-iota 6448  df-fv 6500  df-ov 7363
This theorem is referenced by:  prsthinc  49951
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