Theorem dfnelbr2 44198

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 5672	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}) = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
2		df-xp 5531	. . 3 ⊢ (V × V) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)}
3		df-eprel 5436	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
4	2, 3	difeq12i 4027	. 2 ⊢ ((V × V) ∖ E ) = ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)} ∖ {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦})
5		df-nelbr 44197	. . 3 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦}
6		vex 3414	. . . . . 6 ⊢ 𝑥 ∈ V
7		vex 3414	. . . . . 6 ⊢ 𝑦 ∈ V
8	6, 7	pm3.2i 475	. . . . 5 ⊢ (𝑥 ∈ V ∧ 𝑦 ∈ V)
9	8	biantrur 535	. . . 4 ⊢ (¬ 𝑥 ∈ 𝑦 ↔ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦))
10	9	opabbii 5100	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦} = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
11	5, 10	eqtri 2782	. 2 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ V ∧ 𝑦 ∈ V) ∧ ¬ 𝑥 ∈ 𝑦)}
12	1, 4, 11	3eqtr4ri 2793	1 ⊢ _∉ = ((V × V) ∖ E )

Syntax hints:  ¬ wn 3   ∧ wa 400   = wceq 1539   ∈ wcel 2112  Vcvv 3410   ∖ cdif 3856  {copab 5095   E cep 5435   × cxp 5523   _∉ cnelbr 44196

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2114  ax-9 2122  ax-10 2143  ax-11 2159  ax-12 2176  ax-ext 2730  ax-sep 5170  ax-nul 5177  ax-pr 5299

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 846  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2071  df-clab 2737  df-cleq 2751  df-clel 2831  df-nfc 2902  df-ne 2953  df-ral 3076  df-rex 3077  df-rab 3080  df-v 3412  df-sbc 3698  df-dif 3862  df-un 3864  df-in 3866  df-ss 3876  df-nul 4227  df-if 4422  df-sn 4524  df-pr 4526  df-op 4530  df-opab 5096  df-eprel 5436  df-xp 5531  df-rel 5532  df-nelbr 44197

This theorem is referenced by: (None)