Metamath Proof Explorer


Theorem coeq2d

Description: Equality deduction for composition of two classes. (Contributed by NM, 16-Nov-2000)

Ref Expression
Hypothesis coeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion coeq2d ( 𝜑 → ( 𝐶 ∘ 𝐴 ) = ( 𝐶 ∘ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 coeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 coeq2 ⊢ ( 𝐴 = 𝐵 → ( 𝐶 ∘ 𝐴 ) = ( 𝐶 ∘ 𝐵 ) )
3 1 2 syl ⊢ ( 𝜑 → ( 𝐶 ∘ 𝐴 ) = ( 𝐶 ∘ 𝐵 ) )