Metamath Proof Explorer
Description: Deduction that substitutes equal classes into membership. (Contributed by NM, 14-Dec-2004)
|
|
Ref |
Expression |
|
Hypotheses |
eqeltrrd.1 |
|- ( ph -> A = B ) |
|
|
eqeltrrd.2 |
|- ( ph -> A e. C ) |
|
Assertion |
eqeltrrd |
|- ( ph -> B e. C ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
eqeltrrd.1 |
|- ( ph -> A = B ) |
2 |
|
eqeltrrd.2 |
|- ( ph -> A e. C ) |
3 |
1
|
eqcomd |
|- ( ph -> B = A ) |
4 |
3 2
|
eqeltrd |
|- ( ph -> B e. C ) |