Theorem nelbr 48559

Description: The binary relation of a set not being a member of another set. (Contributed by AV, 26-Dec-2021.)

Assertion

Ref	Expression
nelbr	⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴 _∉ 𝐵 ↔ ¬ 𝐴 ∈ 𝐵))

Proof of Theorem nelbr

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq12 2854	. . 3 ⊢ ((𝑥 = 𝐴 ∧ 𝑦 = 𝐵) → (𝑥 ∈ 𝑦 ↔ 𝐴 ∈ 𝐵))
2	1	notbid 321	. 2 ⊢ ((𝑥 = 𝐴 ∧ 𝑦 = 𝐵) → (¬ 𝑥 ∈ 𝑦 ↔ ¬ 𝐴 ∈ 𝐵))
3		df-nelbr 48557	. 2 ⊢ _∉ = {⟨𝑥, 𝑦⟩ ∣ ¬ 𝑥 ∈ 𝑦}
4	2, 3	brabga 5513	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴 _∉ 𝐵 ↔ ¬ 𝐴 ∈ 𝐵))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2147   class class class wbr 5106   _∉ cnelbr 48556

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-ext 2735  ax-sep 5253  ax-pr 5396

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-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 4839  df-br 5107  df-opab 5172  df-nelbr 48557

This theorem is used by:  nelbrim  48560  nelbrnel  48561