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 ) |