Metamath Proof Explorer

Theorem eceq1d

Description: Equality theorem for equivalence class (deduction form). (Contributed by Jim Kingdon, 31-Dec-2019)

Ref Expression
Hypothesis eceq1d.1 
|- ( ph -> A = B )
Assertion eceq1d 
|- ( ph -> [ A ] C = [ B ] C )

Proof

Step Hyp Ref Expression
1 eceq1d.1 
 |-  ( ph -> A = B )
2 eceq1 
 |-  ( A = B -> [ A ] C = [ B ] C )
3 1 2 syl 
 |-  ( ph -> [ A ] C = [ B ] C )