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Theorem fvconstr2 49105
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 4294 . . 3 (𝜑 → (𝐴𝐹𝐵) ≠ ∅)
3 fvconstr.1 . . . . . . 7 (𝜑𝐹 = (𝑅 × {𝑌}))
43oveqd 7375 . . . . . 6 (𝜑 → (𝐴𝐹𝐵) = (𝐴(𝑅 × {𝑌})𝐵))
5 df-ov 7361 . . . . . 6 (𝐴(𝑅 × {𝑌})𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩)
64, 5eqtrdi 2787 . . . . 5 (𝜑 → (𝐴𝐹𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩))
76neeq1d 2991 . . . 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 2959 . . . . 5 (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}))
118, 10sselid 3931 . . . 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 5099 . 2 (𝐴𝑅𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑅)
1513, 14sylibr 234 1 (𝜑𝐴𝑅𝐵)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1541  wcel 2113  wne 2932  c0 4285  {csn 4580  cop 4586   class class class wbr 5098   × cxp 5622  dom cdm 5624  cfv 6492  (class class class)co 7358
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 2115  ax-9 2123  ax-ext 2708  ax-sep 5241  ax-nul 5251  ax-pr 5377
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 2539  df-eu 2569  df-clab 2715  df-cleq 2728  df-clel 2811  df-ne 2933  df-ral 3052  df-rex 3061  df-rab 3400  df-v 3442  df-dif 3904  df-un 3906  df-ss 3918  df-nul 4286  df-if 4480  df-sn 4581  df-pr 4583  df-op 4587  df-uni 4864  df-br 5099  df-opab 5161  df-xp 5630  df-dm 5634  df-iota 6448  df-fv 6500  df-ov 7361
This theorem is referenced by:  prsthinc  49705
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