Theorem coeq2 4512

Description: Equality theorem for composition of two classes. (Contributed by NM, 3-Jan-1997.)

Assertion

Ref	Expression
coeq2	⊢ (𝐴 = 𝐵 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))

Proof of Theorem coeq2

Step	Hyp	Ref	Expression
1		coss2 4510	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵))
2		coss2 4510	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴))
3	1, 2	anim12i 331	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵) ∧ (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴)))
4		eqss 3014	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3014	. 2 ⊢ ((𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵) ↔ ((𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵) ∧ (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴)))
6	3, 4, 5	3imtr4i 199	1 ⊢ (𝐴 = 𝐵 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))

Syntax hints:   → wi 4   ∧ wa 102   = wceq 1284   ⊆ wss 2973   ∘ ccom 4367

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-in 2979  df-ss 2986  df-br 3786  df-opab 3840  df-co 4372

This theorem is referenced by:  coeq2i  4514  coeq2d  4516  coi2  4857  relcnvtr  4860  relcoi1  4869  f1eqcocnv  5451  ereq1  6136