Theorem dfnelbr2 48616

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 5664	. . 3 ⊢ (V × V) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)}
3		df-eprel 5558	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
4	2, 3	difeq12i 4064	. 2 ⊢ ((V × V) ∖ E ) = ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦})
5		df-nelbr 48615	. . 3 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦}
6		vex 3458	. . . . . 6 ⊢ 𝑥 ∈ V
7		vex 3458	. . . . . 6 ⊢ 𝑦 ∈ V
8	6, 7	pm3.2i 476	. . . . 5 ⊢ (𝑥 ∈ V ∧ 𝑦 ∈ V)
9	8	biantrur 540	. . . 4 ⊢ (¬ 𝑥 ∈ 𝑦 ↔ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦))
10	9	opabbii 5178	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦} = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
11	5, 10	eqtri 2786	. 2 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
12	1, 4, 11	3eqtr4ri 2797	1 ⊢ _∉ = ((V × V) ∖ E )

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 401   = wceq 1570   ∈ wcel 2147  Vcvv 3454   ∖ cdif 4048  {copab 5173   E cep 5557   × cxp 5656   _∉ cnelbr 48614

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-12 2216  ax-ext 2735  ax-sep 5255  ax-pr 5398

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-opab 5174  df-eprel 5558  df-xp 5664  df-rel 5665  df-nelbr 48615

This theorem is used by: (None)