Metamath Proof Explorer
Description: Classes are proper subclasses if and only if their power classes are
proper subclasses. (Contributed by Steven Nguyen, 17-Jul-2022)
|
|
Ref |
Expression |
|
Assertion |
psspwb |
⊢ ( 𝐴 ⊊ 𝐵 ↔ 𝒫 𝐴 ⊊ 𝒫 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
sspwb |
⊢ ( 𝐴 ⊆ 𝐵 ↔ 𝒫 𝐴 ⊆ 𝒫 𝐵 ) |
2 |
|
pweqb |
⊢ ( 𝐴 = 𝐵 ↔ 𝒫 𝐴 = 𝒫 𝐵 ) |
3 |
2
|
necon3bii |
⊢ ( 𝐴 ≠ 𝐵 ↔ 𝒫 𝐴 ≠ 𝒫 𝐵 ) |
4 |
1 3
|
anbi12i |
⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐴 ≠ 𝐵 ) ↔ ( 𝒫 𝐴 ⊆ 𝒫 𝐵 ∧ 𝒫 𝐴 ≠ 𝒫 𝐵 ) ) |
5 |
|
df-pss |
⊢ ( 𝐴 ⊊ 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝐴 ≠ 𝐵 ) ) |
6 |
|
df-pss |
⊢ ( 𝒫 𝐴 ⊊ 𝒫 𝐵 ↔ ( 𝒫 𝐴 ⊆ 𝒫 𝐵 ∧ 𝒫 𝐴 ≠ 𝒫 𝐵 ) ) |
7 |
4 5 6
|
3bitr4i |
⊢ ( 𝐴 ⊊ 𝐵 ↔ 𝒫 𝐴 ⊊ 𝒫 𝐵 ) |