Metamath Proof Explorer
Description: The intersection of a nonempty class exists. Exercise 5 of
TakeutiZaring p. 44 and its converse. (Contributed by NM, 13-Aug-2002)
|
|
Ref |
Expression |
|
Assertion |
intex |
⊢ ( 𝐴 ≠ ∅ ↔ ∩ 𝐴 ∈ V ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
n0 |
⊢ ( 𝐴 ≠ ∅ ↔ ∃ 𝑥 𝑥 ∈ 𝐴 ) |
| 2 |
|
intss1 |
⊢ ( 𝑥 ∈ 𝐴 → ∩ 𝐴 ⊆ 𝑥 ) |
| 3 |
|
vex |
⊢ 𝑥 ∈ V |
| 4 |
3
|
ssex |
⊢ ( ∩ 𝐴 ⊆ 𝑥 → ∩ 𝐴 ∈ V ) |
| 5 |
2 4
|
syl |
⊢ ( 𝑥 ∈ 𝐴 → ∩ 𝐴 ∈ V ) |
| 6 |
5
|
exlimiv |
⊢ ( ∃ 𝑥 𝑥 ∈ 𝐴 → ∩ 𝐴 ∈ V ) |
| 7 |
1 6
|
sylbi |
⊢ ( 𝐴 ≠ ∅ → ∩ 𝐴 ∈ V ) |
| 8 |
|
vprc |
⊢ ¬ V ∈ V |
| 9 |
|
inteq |
⊢ ( 𝐴 = ∅ → ∩ 𝐴 = ∩ ∅ ) |
| 10 |
|
int0 |
⊢ ∩ ∅ = V |
| 11 |
9 10
|
eqtrdi |
⊢ ( 𝐴 = ∅ → ∩ 𝐴 = V ) |
| 12 |
11
|
eleq1d |
⊢ ( 𝐴 = ∅ → ( ∩ 𝐴 ∈ V ↔ V ∈ V ) ) |
| 13 |
8 12
|
mtbiri |
⊢ ( 𝐴 = ∅ → ¬ ∩ 𝐴 ∈ V ) |
| 14 |
13
|
necon2ai |
⊢ ( ∩ 𝐴 ∈ V → 𝐴 ≠ ∅ ) |
| 15 |
7 14
|
impbii |
⊢ ( 𝐴 ≠ ∅ ↔ ∩ 𝐴 ∈ V ) |