Metamath Proof Explorer
Description: Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004)
|
|
Ref |
Expression |
|
Hypotheses |
eqsstrrd.1 |
|- ( ph -> B = A ) |
|
|
eqsstrrd.2 |
|- ( ph -> B C_ C ) |
|
Assertion |
eqsstrrd |
|- ( ph -> A C_ C ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
eqsstrrd.1 |
|- ( ph -> B = A ) |
| 2 |
|
eqsstrrd.2 |
|- ( ph -> B C_ C ) |
| 3 |
1
|
eqcomd |
|- ( ph -> A = B ) |
| 4 |
3 2
|
eqsstrd |
|- ( ph -> A C_ C ) |