Theorem coeq2 4494

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 4492	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵))
2		coss2 4492	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴))
3	1, 2	anim12i 321	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵) ∧ (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴)))
4		eqss 2960	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		eqss 2960	. 2 ⊢ ((𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵) ↔ ((𝐶 ∘ 𝐴) ⊆ (𝐶 ∘ 𝐵) ∧ (𝐶 ∘ 𝐵) ⊆ (𝐶 ∘ 𝐴)))
6	3, 4, 5	3imtr4i 190	1 ⊢ (𝐴 = 𝐵 → (𝐶 ∘ 𝐴) = (𝐶 ∘ 𝐵))

Syntax hints:   → wi 4   ∧ wa 97   = wceq 1243   ⊆ wss 2917   ∘ ccom 4349

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-in 2924  df-ss 2931  df-br 3765  df-opab 3819  df-co 4354

This theorem is referenced by:  coeq2i  4496  coeq2d  4498  coi2  4837  relcnvtr  4840  relcoi1  4849  f1eqcocnv  5431  ereq1  6113