Theorem coeq2 4938

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 4936	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵))
2		coss2 4936	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴))
3	1, 2	anim12i 338	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵) ∧ (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴)))
4		eqss 3263	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3263	. 2 ⊢ ((𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵) ↔ ((𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵) ∧ (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴)))
6	3, 4, 5	3imtr4i 201	1 ⊢ (𝐴 = 𝐵 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   = wceq 1402   ⊆ wss 3220   ∘ ccom 4778

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-in 3226  df-ss 3233  df-br 4131  df-opab 4193  df-co 4783

This theorem is used by:  coeq2i  4940  coeq2d  4942  coi2  5304  relcnvtr  5307  relcoi1  5319  f1eqcocnv  5997  ereq1  6814  seqf1oglem2  10957  seqf1og  10958  gzsumwmhm  13803  gsumvalfi  14152  upxp  15373  uptx  15375  txcn  15376