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 |
|