Metamath Proof Explorer
Description: Equality deduction from subclass relationship and membership.
(Contributed by AV, 21-Aug-2022)
|
|
Ref |
Expression |
|
Hypotheses |
eqelssd.1 |
|
|
|
eqelssd.2 |
|
|
Assertion |
eqelssd |
|
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
eqelssd.1 |
|
| 2 |
|
eqelssd.2 |
|
| 3 |
2
|
ex |
|
| 4 |
3
|
ssrdv |
|
| 5 |
1 4
|
eqssd |
|