Theorem coeq1 4834

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 4832	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶))
2		coss1 4832	. . 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 3207	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3207	. 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 1372   ⊆ wss 3165   ∘ ccom 4678

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 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-nfc 2336  df-in 3171  df-ss 3178  df-br 4044  df-opab 4105  df-co 4683

This theorem is referenced by:  coeq1i  4836  coeq1d  4838  coi2  5198  relcnvtr  5201  funcoeqres  5552  ereq1  6626  updjud  7183  seqf1oglem2  10663  seqf1og  10664