Theorem dfnelbr2 47898

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 5818	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}) = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
2		df-xp 5668	. . 3 ⊢ (V × V) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)}
3		df-eprel 5562	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
4	2, 3	difeq12i 4087	. 2 ⊢ ((V × V) ∖ E ) = ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦})
5		df-nelbr 47897	. . 3 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦}
6		vex 3467	. . . . . 6 ⊢ 𝑥 ∈ V
7		vex 3467	. . . . . 6 ⊢ 𝑦 ∈ V
8	6, 7	pm3.2i 475	. . . . 5 ⊢ (𝑥 ∈ V ∧ 𝑦 ∈ V)
9	8	biantrur 539	. . . 4 ⊢ (¬ 𝑥 ∈ 𝑦 ↔ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦))
10	9	opabbii 5182	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦} = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
11	5, 10	eqtri 2792	. 2 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
12	1, 4, 11	3eqtr4ri 2803	1 ⊢ _∉ = ((V × V) ∖ E )

Syntax hints:  ¬ wn 3   ∧ wa 400   = wceq 1567   ∈ wcel 2149  Vcvv 3463   ∖ cdif 3910  {copab 5177   E cep 5561   × cxp 5660   _∉ cnelbr 47896

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-10 2182  ax-12 2219  ax-ext 2741  ax-sep 5261  ax-pr 5405

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-nf 1811  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-rab 3424  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-opab 5178  df-eprel 5562  df-xp 5668  df-rel 5669  df-nelbr 47897

This theorem is referenced by: (None)