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Theorem fvconstr2 49361
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 4277 . . 3 (𝜑 → (𝐴𝐹𝐵) ≠ ∅)
3 fvconstr.1 . . . . . . 7 (𝜑𝐹 = (𝑅 × {𝑌}))
43oveqd 7380 . . . . . 6 (𝜑 → (𝐴𝐹𝐵) = (𝐴(𝑅 × {𝑌})𝐵))
5 df-ov 7366 . . . . . 6 (𝐴(𝑅 × {𝑌})𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩)
64, 5eqtrdi 2791 . . . . 5 (𝜑 → (𝐴𝐹𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩))
76neeq1d 2994 . . . 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 2962 . . . . 5 (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}))
118, 10sselid 3920 . . . 4 (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
127, 11biimtrdi 254 . . 3 (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅))
132, 12mpd 15 . 2 (𝜑 → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
14 df-br 5080 . 2 (𝐴𝑅𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑅)
1513, 14sylibr 235 1 (𝜑𝐴𝑅𝐵)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1547  wcel 2119  wne 2935  c0 4268  {csn 4562  cop 4568   class class class wbr 5079   × cxp 5623  dom cdm 5625  cfv 6492  (class class class)co 7363
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712  ax-sep 5225  ax-nul 5235  ax-pr 5369
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-mo 2543  df-eu 2573  df-clab 2719  df-cleq 2732  df-clel 2815  df-ne 2936  df-ral 3055  df-rex 3065  df-rab 3393  df-v 3434  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4269  df-if 4462  df-sn 4563  df-pr 4565  df-op 4569  df-uni 4846  df-br 5080  df-opab 5142  df-xp 5631  df-dm 5635  df-iota 6448  df-fv 6500  df-ov 7366
This theorem is referenced by:  prsthinc  49961
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