Metamath Proof Explorer
Description: A deduction from three chained equalities. (Contributed by NM, 4Aug1995) (Proof shortened by Andrew Salmon, 25May2011)


Ref 
Expression 

Hypotheses 
3eqtr3d.1 
 ( ph > A = B ) 


3eqtr3d.2 
 ( ph > A = C ) 


3eqtr3d.3 
 ( ph > B = D ) 

Assertion 
3eqtr3d 
 ( ph > C = D ) 
Proof
Step 
Hyp 
Ref 
Expression 
1 

3eqtr3d.1 
 ( ph > A = B ) 
2 

3eqtr3d.2 
 ( ph > A = C ) 
3 

3eqtr3d.3 
 ( ph > B = D ) 
4 
1 2

eqtr3d 
 ( ph > B = C ) 
5 
4 3

eqtr3d 
 ( ph > C = D ) 