Theorem coeq1 4542

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

Assertion

Ref	Expression
coeq1	⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐶) = (𝐵 ∘ 𝐶))

Proof of Theorem coeq1

Step	Hyp	Ref	Expression
1		coss1 4540	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶))
2		coss1 4540	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝐵 ∘ 𝐶) ⊆ (𝐴 ∘ 𝐶))
3	1, 2	anim12i 331	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶) ∧ (𝐵 ∘ 𝐶) ⊆ (𝐴 ∘ 𝐶)))
4		eqss 3024	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3024	. 2 ⊢ ((𝐴 ∘ 𝐶) = (𝐵 ∘ 𝐶) ↔ ((𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶) ∧ (𝐵 ∘ 𝐶) ⊆ (𝐴 ∘ 𝐶)))
6	3, 4, 5	3imtr4i 199	1 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐶) = (𝐵 ∘ 𝐶))

Syntax hints:   → wi 4   ∧ wa 102   = wceq 1285   ⊆ wss 2983   ∘ ccom 4396

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 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-in 2989  df-ss 2996  df-br 3807  df-opab 3861  df-co 4401

This theorem is referenced by:  coeq1i  4544  coeq1d  4546  coi2  4888  relcnvtr  4891  funcoeqres  5209  ereq1  6201