Theorem coeq2d 4789

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 4785	. 2 ⊢ (𝐴 = 𝐵 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))

Syntax hints:   → wi 4   = wceq 1353   ∘ ccom 4630

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-in 3135  df-ss 3142  df-br 4004  df-opab 4065  df-co 4635

This theorem is referenced by:  coeq12d  4791  relcoi1  5160  f1ococnv1  5490  funcoeqres  5492  fcof1o  5789  foeqcnvco  5790  mapen  6845  hashfacen  10811