Theorem nelb 3238

Description: A definition of ¬ 𝐴 ∈ 𝐵. (Contributed by Thierry Arnoux, 20-Nov-2023.) (Proof shortened by SN, 23-Jan-2024.) (Proof shortened by Wolf Lammen, 3-Nov-2024.)

Assertion

Ref	Expression
nelb	⊢ (¬ 𝐴 ∈ 𝐵 ↔ ∀𝑥 ∈ 𝐵 𝑥 ≠ 𝐴)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nelb

Step	Hyp	Ref	Expression
1		df-ne 2956	. . . 4 ⊢ (𝑥 ≠ 𝐴 ↔ ¬ 𝑥 = 𝐴)
2	1	ralbii 3108	. . 3 ⊢ (∀𝑥 ∈ 𝐵 𝑥 ≠ 𝐴 ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 = 𝐴)
3		ralnex 3088	. . 3 ⊢ (∀𝑥 ∈ 𝐵 ¬ 𝑥 = 𝐴 ↔ ¬ ∃𝑥 ∈ 𝐵 𝑥 = 𝐴)
4	2, 3	bitr2i 279	. 2 ⊢ (¬ ∃𝑥 ∈ 𝐵 𝑥 = 𝐴 ↔ ∀𝑥 ∈ 𝐵 𝑥 ≠ 𝐴)
5		risset 3237	. 2 ⊢ (𝐴 ∈ 𝐵 ↔ ∃𝑥 ∈ 𝐵 𝑥 = 𝐴)
6	4, 5	xchnxbir 336	1 ⊢ (¬ 𝐴 ∈ 𝐵 ↔ ∀𝑥 ∈ 𝐵 𝑥 ≠ 𝐴)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   = wceq 1569   ∈ wcel 2144   ≠ wne 2955  ∀wral 3076  ∃wrex 3086

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-clel 2835  df-ne 2956  df-ral 3077  df-rex 3087

This theorem is used by:  dfdif3  4055  inpr0  33128