Theorem coeq1 4768

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 4766	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶))
2		coss1 4766	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝐵 ∘ 𝐶) ⊆ (𝐴 ∘ 𝐶))
3	1, 2	anim12i 336	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶) ∧ (𝐵 ∘ 𝐶) ⊆ (𝐴 ∘ 𝐶)))
4		eqss 3162	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 3162	. 2 ⊢ ((𝐴 ∘ 𝐶) = (𝐵 ∘ 𝐶) ↔ ((𝐴 ∘ 𝐶) ⊆ (𝐵 ∘ 𝐶) ∧ (𝐵 ∘ 𝐶) ⊆ (𝐴 ∘ 𝐶)))
6	3, 4, 5	3imtr4i 200	1 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐶) = (𝐵 ∘ 𝐶))

Syntax hints:   → wi 4   ∧ wa 103   = wceq 1348   ⊆ wss 3121   ∘ ccom 4615

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 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-in 3127  df-ss 3134  df-br 3990  df-opab 4051  df-co 4620

This theorem is referenced by:  coeq1i  4770  coeq1d  4772  coi2  5127  relcnvtr  5130  funcoeqres  5473  ereq1  6520  updjud  7059