Theorem coeq2d 4498

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

Hypothesis

Ref	Expression
coeq1d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
coeq2d	⊢ (𝜑 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))

Proof of Theorem coeq2d

Step	Hyp	Ref	Expression
1		coeq1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		coeq2 4494	. 2 ⊢ (𝐴 = 𝐵 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))

Syntax hints:   → wi 4   = wceq 1243   ∘ ccom 4349

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-in 2924  df-ss 2931  df-br 3765  df-opab 3819  df-co 4354

This theorem is referenced by:  coeq12d  4500  relcoi1  4849  f1ococnv1  5155  funcoeqres  5157  fcof1o  5429  foeqcnvco  5430