Theorem dfnelbr2 47867

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 5803	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}) = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
2		df-xp 5653	. . 3 ⊢ (V × V) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)}
3		df-eprel 5547	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
4	2, 3	difeq12i 4078	. 2 ⊢ ((V × V) ∖ E ) = ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦})
5		df-nelbr 47866	. . 3 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦}
6		vex 3458	. . . . . 6 ⊢ 𝑥 ∈ V
7		vex 3458	. . . . . 6 ⊢ 𝑦 ∈ V
8	6, 7	pm3.2i 474	. . . . 5 ⊢ (𝑥 ∈ V ∧ 𝑦 ∈ V)
9	8	biantrur 538	. . . 4 ⊢ (¬ 𝑥 ∈ 𝑦 ↔ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦))
10	9	opabbii 5167	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦} = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
11	5, 10	eqtri 2785	. 2 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
12	1, 4, 11	3eqtr4ri 2796	1 ⊢ _∉ = ((V × V) ∖ E )

Syntax hints:  ¬ wn 3   ∧ wa 399   = wceq 1560   ∈ wcel 2142  Vcvv 3454   ∖ cdif 3901  {copab 5162   E cep 5546   × cxp 5645   _∉ cnelbr 47865

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-10 2175  ax-12 2212  ax-ext 2734  ax-sep 5246  ax-pr 5390

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1100  df-tru 1563  df-fal 1573  df-ex 1800  df-nf 1804  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4481  df-sn 4583  df-pr 4585  df-op 4589  df-opab 5163  df-eprel 5547  df-xp 5653  df-rel 5654  df-nelbr 47866

This theorem is referenced by: (None)