Theorem fvconstr2 49217

Description: Two ways of expressing 𝐴𝑅𝐵. (Contributed by Zhi Wang, 18-Sep-2024.)

Hypotheses

Ref	Expression
fvconstr.1	⊢ (𝜑 → 𝐹 = (𝑅 × {𝑌}))
fvconstr2.2	⊢ (𝜑 → 𝑋 ∈ (𝐴𝐹𝐵))

Assertion

Ref	Expression
fvconstr2	⊢ (𝜑 → 𝐴𝑅𝐵)

Proof of Theorem fvconstr2

Step	Hyp	Ref	Expression
1		fvconstr2.2	. . . 4 ⊢ (𝜑 → 𝑋 ∈ (𝐴𝐹𝐵))
2	1	ne0d 4296	. . 3 ⊢ (𝜑 → (𝐴𝐹𝐵) ≠ ∅)
3		fvconstr.1	. . . . . . 7 ⊢ (𝜑 → 𝐹 = (𝑅 × {𝑌}))
4	3	oveqd 7385	. . . . . 6 ⊢ (𝜑 → (𝐴𝐹𝐵) = (𝐴(𝑅 × {𝑌})𝐵))
5		df-ov 7371	. . . . . 6 ⊢ (𝐴(𝑅 × {𝑌})𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩)
6	4, 5	eqtrdi 2788	. . . . 5 ⊢ (𝜑 → (𝐴𝐹𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩))
7	6	neeq1d 2992	. . . 4 ⊢ (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ ↔ ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅))
8		dmxpss 6137	. . . . 5 ⊢ dom (𝑅 × {𝑌}) ⊆ 𝑅
9		ndmfv 6874	. . . . . 6 ⊢ (¬ ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}) → ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) = ∅)
10	9	necon1ai 2960	. . . . 5 ⊢ (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}))
11	8, 10	sselid 3933	. . . 4 ⊢ (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
12	7, 11	biimtrdi 253	. . 3 ⊢ (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅))
13	2, 12	mpd 15	. 2 ⊢ (𝜑 → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
14		df-br 5101	. 2 ⊢ (𝐴𝑅𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑅)
15	13, 14	sylibr 234	1 ⊢ (𝜑 → 𝐴𝑅𝐵)

Syntax hints:   → wi 4   = wceq 1542   ∈ wcel 2114   ≠ wne 2933  ∅c0 4287  {csn 4582  ⟨cop 4588   class class class wbr 5100   × cxp 5630  dom cdm 5632  ‘cfv 6500  (class class class)co 7368

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 5243  ax-nul 5253  ax-pr 5379

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 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-uni 4866  df-br 5101  df-opab 5163  df-xp 5638  df-dm 5642  df-iota 6456  df-fv 6508  df-ov 7371

This theorem is referenced by:  prsthinc  49817