Theorem nelbOLDOLD

Description: Obsolete version of nelb as of 23-Jan-2024. (Contributed by Thierry Arnoux, 20-Nov-2023) (New usage is discouraged.) (Proof modification is discouraged.)

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

Proof

Step	Hyp	Ref	Expression
1		df-ral	⊢ ( ∀ 𝑥 ∈ 𝐵 ¬ 𝑥 = 𝐴 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ¬ 𝑥 = 𝐴 ) )
2		df-ne	⊢ ( 𝑥 ≠ 𝐴 ↔ ¬ 𝑥 = 𝐴 )
3	2	ralbii	⊢ ( ∀ 𝑥 ∈ 𝐵 𝑥 ≠ 𝐴 ↔ ∀ 𝑥 ∈ 𝐵 ¬ 𝑥 = 𝐴 )
4		dfclel	⊢ ( 𝐴 ∈ 𝐵 ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐵 ) )
5	4	notbii	⊢ ( ¬ 𝐴 ∈ 𝐵 ↔ ¬ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐵 ) )
6		alnex	⊢ ( ∀ 𝑥 ¬ ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐵 ) ↔ ¬ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐵 ) )
7		imnan	⊢ ( ( 𝑥 ∈ 𝐵 → ¬ 𝑥 = 𝐴 ) ↔ ¬ ( 𝑥 ∈ 𝐵 ∧ 𝑥 = 𝐴 ) )
8		ancom	⊢ ( ( 𝑥 ∈ 𝐵 ∧ 𝑥 = 𝐴 ) ↔ ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐵 ) )
9	8	notbii	⊢ ( ¬ ( 𝑥 ∈ 𝐵 ∧ 𝑥 = 𝐴 ) ↔ ¬ ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐵 ) )
10	7 9	bitr2i	⊢ ( ¬ ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐵 → ¬ 𝑥 = 𝐴 ) )
11	10	albii	⊢ ( ∀ 𝑥 ¬ ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ¬ 𝑥 = 𝐴 ) )
12	5 6 11	3bitr2i	⊢ ( ¬ 𝐴 ∈ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ¬ 𝑥 = 𝐴 ) )
13	1 3 12	3bitr4ri	⊢ ( ¬ 𝐴 ∈ 𝐵 ↔ ∀ 𝑥 ∈ 𝐵 𝑥 ≠ 𝐴 )