Theorem fvconstr2 49449

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 4294	. . 3 ⊢ (𝜑 → (𝐴𝐹𝐵) ≠ ∅)
3		fvconstr.1	. . . . . . 7 ⊢ (𝜑 → 𝐹 = (𝑅 × {𝑌}))
4	3	oveqd 7409	. . . . . 6 ⊢ (𝜑 → (𝐴𝐹𝐵) = (𝐴(𝑅 × {𝑌})𝐵))
5		df-ov 7395	. . . . . 6 ⊢ (𝐴(𝑅 × {𝑌})𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩)
6	4, 5	eqtrdi 2812	. . . . 5 ⊢ (𝜑 → (𝐴𝐹𝐵) = ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩))
7	6	neeq1d 3015	. . . 4 ⊢ (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ ↔ ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅))
8		dmxpss 6153	. . . . 5 ⊢ dom (𝑅 × {𝑌}) ⊆ 𝑅
9		ndmfv 6895	. . . . . 6 ⊢ (¬ ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}) → ((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) = ∅)
10	9	necon1ai 2983	. . . . 5 ⊢ (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ dom (𝑅 × {𝑌}))
11	8, 10	sselid 3934	. . . 4 ⊢ (((𝑅 × {𝑌})‘⟨𝐴, 𝐵⟩) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
12	7, 11	biimtrdi 255	. . 3 ⊢ (𝜑 → ((𝐴𝐹𝐵) ≠ ∅ → ⟨𝐴, 𝐵⟩ ∈ 𝑅))
13	2, 12	mpd 15	. 2 ⊢ (𝜑 → ⟨𝐴, 𝐵⟩ ∈ 𝑅)
14		df-br 5100	. 2 ⊢ (𝐴𝑅𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑅)
15	13, 14	sylibr 236	1 ⊢ (𝜑 → 𝐴𝑅𝐵)

Syntax hints:   → wi 4   = wceq 1559   ∈ wcel 2141   ≠ wne 2956  ∅c0 4285  {csn 4581  ⟨cop 4587   class class class wbr 5099   × cxp 5643  dom cdm 5645  ‘cfv 6517  (class class class)co 7392

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733  ax-sep 5245  ax-nul 5255  ax-pr 5389

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1099  df-tru 1562  df-fal 1572  df-ex 1799  df-sb 2090  df-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-ral 3076  df-rex 3086  df-rab 3414  df-v 3455  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4480  df-sn 4582  df-pr 4584  df-op 4588  df-uni 4865  df-br 5100  df-opab 5162  df-xp 5651  df-dm 5655  df-iota 6473  df-fv 6525  df-ov 7395

This theorem is referenced by:  prsthinc  50049