Theorem coeq2 4762

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 4760	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵))
2		coss2 4760	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴))
3	1, 2	anim12i 336	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵) ∧ (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴)))
4		eqss 3157	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3157	. 2 ⊢ ((𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵) ↔ ((𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵) ∧ (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴)))
6	3, 4, 5	3imtr4i 200	1 ⊢ (𝐴 = 𝐵 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))

Syntax hints:   → wi 4   ∧ wa 103   = wceq 1343   ⊆ wss 3116   ∘ ccom 4608

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-in 3122  df-ss 3129  df-br 3983  df-opab 4044  df-co 4613

This theorem is referenced by:  coeq2i  4764  coeq2d  4766  coi2  5120  relcnvtr  5123  relcoi1  5135  f1eqcocnv  5759  ereq1  6508  upxp  12912  uptx  12914  txcn  12915