Metamath Proof Explorer
Description: Deduce subclass from proper subclass. (Contributed by NM, 29-Feb-1996)
|
|
Ref |
Expression |
|
Hypothesis |
pssssd.1 |
|- ( ph -> A C. B ) |
|
Assertion |
pssssd |
|- ( ph -> A C_ B ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
pssssd.1 |
|- ( ph -> A C. B ) |
| 2 |
|
pssss |
|- ( A C. B -> A C_ B ) |
| 3 |
1 2
|
syl |
|- ( ph -> A C_ B ) |