Metamath Proof Explorer


Theorem ralfal

Description: Two ways of expressing empty set. (Contributed by Glauco Siliprandi, 24-Jan-2024)

Ref Expression
Hypothesis ralfal.1 𝑥 𝐴
Assertion ralfal ( 𝐴 = ∅ ↔ ∀ 𝑥𝐴 ⊥ )

Proof

Step Hyp Ref Expression
1 ralfal.1 𝑥 𝐴
2 df-fal ( ⊥ ↔ ¬ ⊤ )
3 2 ralbii ( ∀ 𝑥𝐴 ⊥ ↔ ∀ 𝑥𝐴 ¬ ⊤ )
4 ralnex ( ∀ 𝑥𝐴 ¬ ⊤ ↔ ¬ ∃ 𝑥𝐴 ⊤ )
5 3 4 bitri ( ∀ 𝑥𝐴 ⊥ ↔ ¬ ∃ 𝑥𝐴 ⊤ )
6 rextru ( ∃ 𝑥 𝑥𝐴 ↔ ∃ 𝑥𝐴 ⊤ )
7 6 notbii ( ¬ ∃ 𝑥 𝑥𝐴 ↔ ¬ ∃ 𝑥𝐴 ⊤ )
8 1 neq0f ( ¬ 𝐴 = ∅ ↔ ∃ 𝑥 𝑥𝐴 )
9 8 con1bii ( ¬ ∃ 𝑥 𝑥𝐴𝐴 = ∅ )
10 5 7 9 3bitr2ri ( 𝐴 = ∅ ↔ ∀ 𝑥𝐴 ⊥ )