Metamath Proof Explorer
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 |
⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝑥 = 𝐵 ) |
|
|
eqsnd.2 |
⊢ ( 𝜑 → 𝐵 ∈ 𝐴 ) |
|
Assertion |
eqsnd |
⊢ ( 𝜑 → 𝐴 = { 𝐵 } ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
eqsnd.1 |
⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝑥 = 𝐵 ) |
2 |
|
eqsnd.2 |
⊢ ( 𝜑 → 𝐵 ∈ 𝐴 ) |
3 |
1
|
ralrimiva |
⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝐴 𝑥 = 𝐵 ) |
4 |
2
|
ne0d |
⊢ ( 𝜑 → 𝐴 ≠ ∅ ) |
5 |
|
eqsn |
⊢ ( 𝐴 ≠ ∅ → ( 𝐴 = { 𝐵 } ↔ ∀ 𝑥 ∈ 𝐴 𝑥 = 𝐵 ) ) |
6 |
4 5
|
syl |
⊢ ( 𝜑 → ( 𝐴 = { 𝐵 } ↔ ∀ 𝑥 ∈ 𝐴 𝑥 = 𝐵 ) ) |
7 |
3 6
|
mpbird |
⊢ ( 𝜑 → 𝐴 = { 𝐵 } ) |