Metamath Proof Explorer


Theorem eqsnd

Description: Deduce that a set is a singleton. (Contributed by Thierry Arnoux, 10-May-2023) (Proof shortened by SN, 3-Jul-2025)

Ref Expression
Hypotheses eqsnd.1 ( ( 𝜑𝑥𝐴 ) → 𝑥 = 𝐵 )
eqsnd.2 ( 𝜑𝐵𝐴 )
Assertion eqsnd ( 𝜑𝐴 = { 𝐵 } )

Proof

Step Hyp Ref Expression
1 eqsnd.1 ( ( 𝜑𝑥𝐴 ) → 𝑥 = 𝐵 )
2 eqsnd.2 ( 𝜑𝐵𝐴 )
3 1 ralrimiva ( 𝜑 → ∀ 𝑥𝐴 𝑥 = 𝐵 )
4 2 ne0d ( 𝜑𝐴 ≠ ∅ )
5 eqsn ( 𝐴 ≠ ∅ → ( 𝐴 = { 𝐵 } ↔ ∀ 𝑥𝐴 𝑥 = 𝐵 ) )
6 4 5 syl ( 𝜑 → ( 𝐴 = { 𝐵 } ↔ ∀ 𝑥𝐴 𝑥 = 𝐵 ) )
7 3 6 mpbird ( 𝜑𝐴 = { 𝐵 } )