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 ( 𝜑𝐴 = 𝐵 )
Assertion eceq1d ( 𝜑 → [ 𝐴 ] 𝐶 = [ 𝐵 ] 𝐶 )

Proof

Step Hyp Ref Expression
1 eceq1d.1 ( 𝜑𝐴 = 𝐵 )
2 eceq1 ( 𝐴 = 𝐵 → [ 𝐴 ] 𝐶 = [ 𝐵 ] 𝐶 )
3 1 2 syl ( 𝜑 → [ 𝐴 ] 𝐶 = [ 𝐵 ] 𝐶 )