Metamath Proof Explorer
Description: An equality transitivity deduction. (Contributed by NM, 18Oct1999)


Ref 
Expression 

Hypotheses 
eqtr2d.1 
 ( ph > A = B ) 


eqtr2d.2 
 ( ph > B = C ) 

Assertion 
eqtr2d 
 ( ph > C = A ) 
Proof
Step 
Hyp 
Ref 
Expression 
1 

eqtr2d.1 
 ( ph > A = B ) 
2 

eqtr2d.2 
 ( ph > B = C ) 
3 
1 2

eqtrd 
 ( ph > A = C ) 
4 
3

eqcomd 
 ( ph > C = A ) 