Metamath Proof Explorer
Description: An equality transitivity equality deduction. (Contributed by NM, 18-Jul-1995)
|
|
Ref |
Expression |
|
Hypotheses |
eqtr4d.1 |
|- ( ph -> A = B ) |
|
|
eqtr4d.2 |
|- ( ph -> C = B ) |
|
Assertion |
eqtr4d |
|- ( ph -> A = C ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
eqtr4d.1 |
|- ( ph -> A = B ) |
2 |
|
eqtr4d.2 |
|- ( ph -> C = B ) |
3 |
2
|
eqcomd |
|- ( ph -> B = C ) |
4 |
1 3
|
eqtrd |
|- ( ph -> A = C ) |