Theorem dfnelbr2 44652

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 5729	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}) = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
2		df-xp 5586	. . 3 ⊢ (V × V) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)}
3		df-eprel 5486	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
4	2, 3	difeq12i 4051	. 2 ⊢ ((V × V) ∖ E ) = ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦})
5		df-nelbr 44651	. . 3 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦}
6		vex 3426	. . . . . 6 ⊢ 𝑥 ∈ V
7		vex 3426	. . . . . 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 5137	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦} = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
11	5, 10	eqtri 2766	. 2 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
12	1, 4, 11	3eqtr4ri 2777	1 ⊢ _∉ = ((V × V) ∖ E )

Syntax hints:  ¬ wn 3   ∧ wa 395   = wceq 1539   ∈ wcel 2108  Vcvv 3422   ∖ cdif 3880  {copab 5132   E cep 5485   × cxp 5578   _∉ cnelbr 44650

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-ne 2943  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-sbc 3712  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-opab 5133  df-eprel 5486  df-xp 5586  df-rel 5587  df-nelbr 44651

This theorem is referenced by: (None)