Theorem fvconstr2 49619

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 7429	. . . . . 6 ⊢ (𝜑 → (𝐴𝐹𝐵) = (𝐴(𝑅 × {𝑌})𝐵))
5		df-ov 7415	. . . . . 6 ⊢ (𝐴(𝑅 × {𝑌})𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩)
6	4, 5	eqtrdi 2814	. . . . 5 ⊢ (𝜑 → (𝐴𝐹𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩))
7	6	neeq1d 3017	. . . 4 ⊢ (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ ↔ ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅))
8		dmxpss 6171	. . . . 5 ⊢ dom (𝑅 × {𝑌}) ⊆ 𝑅
9		ndmfv 6915	. . . . . 6 ⊢ (¬ ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}) → ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) = ∅)
10	9	necon1ai 2985	. . . . 5 ⊢ (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}))
11	8, 10	sselid 3936	. . . 4 ⊢ (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
12	7, 11	biimtrdi 256	. . 3 ⊢ (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅))
13	2, 12	mpd 16	. 2 ⊢ (𝜑 → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
14		df-br 5111	. 2 ⊢ (𝐴𝑅𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑅)
15	13, 14	sylibr 237	1 ⊢ (𝜑 → 𝐴𝑅𝐵)

Syntax hints:   → wi 4   = wceq 1570   ∈ wcel 2143   ≠ wne 2958  ∅c0 4287  {csn 4590  ⟨cop 4596   class class class wbr 5110   × cxp 5661  dom cdm 5663  ‘cfv 6538  (class class class)co 7412

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735  ax-sep 5258  ax-nul 5270  ax-pr 5406

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2838  df-ne 2959  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4288  df-if 4489  df-sn 4591  df-pr 4593  df-op 4597  df-uni 4874  df-br 5111  df-opab 5175  df-xp 5669  df-dm 5673  df-iota 6494  df-fv 6546  df-ov 7415

This theorem is referenced by:  prsthinc  50219