Theorem djueq12 6978

Description: Equality theorem for disjoint union. (Contributed by Jim Kingdon, 23-Jun-2022.)

Assertion

Ref	Expression
djueq12	⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 ⊔ 𝐶) = (𝐵 ⊔ 𝐷))

Proof of Theorem djueq12

Step	Hyp	Ref	Expression
1		xpeq2 4600	. . . 4 ⊢ (𝐴 = 𝐵 → ({∅} × 𝐴) = ({∅} × 𝐵))
2	1	adantr 274	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → ({∅} × 𝐴) = ({∅} × 𝐵))
3		xpeq2 4600	. . . 4 ⊢ (𝐶 = 𝐷 → ({1o} × 𝐶) = ({1o} × 𝐷))
4	3	adantl 275	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → ({1o} × 𝐶) = ({1o} × 𝐷))
5	2, 4	uneq12d 3262	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (({∅} × 𝐴) ∪ ({1o} × 𝐶)) = (({∅} × 𝐵) ∪ ({1o} × 𝐷)))
6		df-dju 6977	. 2 ⊢ (𝐴 ⊔ 𝐶) = (({∅} × 𝐴) ∪ ({1o} × 𝐶))
7		df-dju 6977	. 2 ⊢ (𝐵 ⊔ 𝐷) = (({∅} × 𝐵) ∪ ({1o} × 𝐷))
8	5, 6, 7	3eqtr4g 2215	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 ⊔ 𝐶) = (𝐵 ⊔ 𝐷))

Syntax hints:   → wi 4   ∧ wa 103   = wceq 1335   ∪ cun 3100  ∅c0 3394  {csn 3560   × cxp 4583  1_oc1o 6353   ⊔ cdju 6976

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 699  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139

This theorem depends on definitions:  df-bi 116  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-v 2714  df-un 3106  df-opab 4026  df-xp 4591  df-dju 6977

This theorem is referenced by:  djueq1  6979  djueq2  6980  casef  7027