Theorem dfnelbr2 43010

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 5595	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}) = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
2		df-xp 5456	. . 3 ⊢ (V × V) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)}
3		df-eprel 5360	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
4	2, 3	difeq12i 4024	. 2 ⊢ ((V × V) ∖ E ) = ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦})
5		df-nelbr 43009	. . 3 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦}
6		vex 3443	. . . . . 6 ⊢ 𝑥 ∈ V
7		vex 3443	. . . . . 6 ⊢ 𝑦 ∈ V
8	6, 7	pm3.2i 471	. . . . 5 ⊢ (𝑥 ∈ V ∧ 𝑦 ∈ V)
9	8	biantrur 531	. . . 4 ⊢ (¬ 𝑥 ∈ 𝑦 ↔ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦))
10	9	opabbii 5035	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦} = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
11	5, 10	eqtri 2821	. 2 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
12	1, 4, 11	3eqtr4ri 2832	1 ⊢ _∉ = ((V × V) ∖ E )

Syntax hints:  ¬ wn 3   ∧ wa 396   = wceq 1525   ∈ wcel 2083  Vcvv 3440   ∖ cdif 3862  {copab 5030   E cep 5359   × cxp 5448   _∉ cnelbr 43008

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1781  ax-4 1795  ax-5 1892  ax-6 1951  ax-7 1996  ax-8 2085  ax-9 2093  ax-10 2114  ax-11 2128  ax-12 2143  ax-13 2346  ax-ext 2771  ax-sep 5101  ax-nul 5108  ax-pr 5228

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-3an 1082  df-tru 1528  df-ex 1766  df-nf 1770  df-sb 2045  df-mo 2578  df-eu 2614  df-clab 2778  df-cleq 2790  df-clel 2865  df-nfc 2937  df-ne 2987  df-ral 3112  df-rex 3113  df-rab 3116  df-v 3442  df-sbc 3712  df-dif 3868  df-un 3870  df-in 3872  df-ss 3880  df-nul 4218  df-if 4388  df-sn 4479  df-pr 4481  df-op 4485  df-opab 5031  df-eprel 5360  df-xp 5456  df-rel 5457  df-nelbr 43009

This theorem is referenced by: (None)