Theorem coeq1 4887

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 4885	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶))
2		coss1 4885	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝐵 ∘ 𝐶) ⊆ (𝐴 ∘ 𝐶))
3	1, 2	anim12i 338	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶) ∧ (𝐵 ∘ 𝐶) ⊆ (𝐴 ∘ 𝐶)))
4		eqss 3242	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3242	. 2 ⊢ ((𝐴 ∘ 𝐶) = (𝐵 ∘ 𝐶) ↔ ((𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶) ∧ (𝐵 ∘ 𝐶) ⊆ (𝐴 ∘ 𝐶)))
6	3, 4, 5	3imtr4i 201	1 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐶) = (𝐵 ∘ 𝐶))

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1397   ⊆ wss 3200   ∘ ccom 4729

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-in 3206  df-ss 3213  df-br 4089  df-opab 4151  df-co 4734

This theorem is referenced by:  coeq1i  4889  coeq1d  4891  coi2  5253  relcnvtr  5256  funcoeqres  5614  ereq1  6708  updjud  7280  seqf1oglem2  10781  seqf1og  10782  gfsumval  16680