Theorem dfnelbr2 47290

Description: Alternate definition of the negated membership as binary relation. (Proposed by BJ, 27-Dec-2021.) (Contributed by AV, 27-Dec-2021.)

Assertion

Ref	Expression
dfnelbr2	⊢ _∉ = ((V × V) ∖ E )

Proof of Theorem dfnelbr2

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		difopab 5839	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}) = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
2		df-xp 5690	. . 3 ⊢ (V × V) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)}
3		df-eprel 5583	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
4	2, 3	difeq12i 4123	. 2 ⊢ ((V × V) ∖ E ) = ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦})
5		df-nelbr 47289	. . 3 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦}
6		vex 3483	. . . . . 6 ⊢ 𝑥 ∈ V
7		vex 3483	. . . . . 6 ⊢ 𝑦 ∈ V
8	6, 7	pm3.2i 470	. . . . 5 ⊢ (𝑥 ∈ V ∧ 𝑦 ∈ V)
9	8	biantrur 530	. . . 4 ⊢ (¬ 𝑥 ∈ 𝑦 ↔ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦))
10	9	opabbii 5209	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦} = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
11	5, 10	eqtri 2764	. 2 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
12	1, 4, 11	3eqtr4ri 2775	1 ⊢ _∉ = ((V × V) ∖ E )

Syntax hints:  ¬ wn 3   ∧ wa 395   = wceq 1539   ∈ wcel 2107  Vcvv 3479   ∖ cdif 3947  {copab 5204   E cep 5582   × cxp 5682   _∉ cnelbr 47288

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-10 2140  ax-12 2176  ax-ext 2707  ax-sep 5295  ax-nul 5305  ax-pr 5431

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1779  df-nf 1783  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-rab 3436  df-v 3481  df-dif 3953  df-un 3955  df-ss 3967  df-nul 4333  df-if 4525  df-sn 4626  df-pr 4628  df-op 4632  df-opab 5205  df-eprel 5583  df-xp 5690  df-rel 5691  df-nelbr 47289

This theorem is referenced by: (None)