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Theorem fvconstr2 49683
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 4298 . . 3 (𝜑 → (𝐴𝐹𝐵) ≠ ∅)
3 fvconstr.1 . . . . . . 7 (𝜑𝐹 = (𝑅 × {𝑌}))
43oveqd 7440 . . . . . 6 (𝜑 → (𝐴𝐹𝐵) = (𝐴(𝑅 × {𝑌})𝐵))
5 df-ov 7426 . . . . . 6 (𝐴(𝑅 × {𝑌})𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩)
64, 5eqtrdi 2817 . . . . 5 (𝜑 → (𝐴𝐹𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩))
76neeq1d 3020 . . . 4 (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ ↔ ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅))
8 dmxpss 6174 . . . . 5 dom (𝑅 × {𝑌}) ⊆ 𝑅
9 ndmfv 6920 . . . . . 6 (¬ ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}) → ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) = ∅)
109necon1ai 2988 . . . . 5 (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}))
118, 10sselid 3938 . . . 4 (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
127, 11biimtrdi 256 . . 3 (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅))
132, 12mpd 16 . 2 (𝜑 → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
14 df-br 5115 . 2 (𝐴𝑅𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑅)
1513, 14sylibr 237 1 (𝜑𝐴𝑅𝐵)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4   = wceq 1570  wcel 2146  wne 2961  c0 4289  {csn 4594  cop 4600   class class class wbr 5114   × cxp 5664  dom cdm 5666  cfv 6543  (class class class)co 7423
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738  ax-sep 5262  ax-nul 5274  ax-pr 5409
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-mo 2570  df-eu 2600  df-clab 2745  df-cleq 2758  df-clel 2841  df-ne 2962  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-uni 4878  df-br 5115  df-opab 5179  df-xp 5672  df-dm 5676  df-iota 6499  df-fv 6551  df-ov 7426
This theorem is used by:  prsthinc  50283
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