Theorem uneqdifeq 4448

Description: Two ways to say that 𝐴 and 𝐵 partition 𝐶 (when 𝐴 and 𝐵 don't overlap and 𝐴 is a part of 𝐶). (Contributed by FL, 17-Nov-2008.) (Proof shortened by JJ, 14-Jul-2021.)

Assertion

Ref	Expression
uneqdifeq	⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐴 ∪ 𝐵) = 𝐶 ↔ (𝐶 ∖ 𝐴) = 𝐵))

Proof of Theorem uneqdifeq

Step	Hyp	Ref	Expression
1		uncom 4113	. . . . 5 ⊢ (𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵)
2		eqtr 2784	. . . . . . 7 ⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → (𝐵 ∪ 𝐴) = 𝐶)
3	2	eqcomd 2770	. . . . . 6 ⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → 𝐶 = (𝐵 ∪ 𝐴))
4		difeq1 4075	. . . . . . 7 ⊢ (𝐶 = (𝐵 ∪ 𝐴) → (𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴))
5		difun2 4437	. . . . . . 7 ⊢ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)
6		eqtr 2784	. . . . . . . 8 ⊢ (((𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴) ∧ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)) → (𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴))
7		ineqcom 4164	. . . . . . . . . 10 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅)
8		disj3 4410	. . . . . . . . . 10 ⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ 𝐵 = (𝐵 ∖ 𝐴))
9	7, 8	bitri 277	. . . . . . . . 9 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐵 = (𝐵 ∖ 𝐴))
10		eqtr 2784	. . . . . . . . . . 11 ⊢ (((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) ∧ (𝐵 ∖ 𝐴) = 𝐵) → (𝐶 ∖ 𝐴) = 𝐵)
11	10	expcom 417	. . . . . . . . . 10 ⊢ ((𝐵 ∖ 𝐴) = 𝐵 → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵))
12	11	eqcoms 2772	. . . . . . . . 9 ⊢ (𝐵 = (𝐵 ∖ 𝐴) → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵))
13	9, 12	sylbi 219	. . . . . . . 8 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵))
14	6, 13	syl5com 31	. . . . . . 7 ⊢ (((𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴) ∧ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
15	4, 5, 14	sylancl 595	. . . . . 6 ⊢ (𝐶 = (𝐵 ∪ 𝐴) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
16	3, 15	syl 17	. . . . 5 ⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
17	1, 16	mpan 700	. . . 4 ⊢ ((𝐴 ∪ 𝐵) = 𝐶 → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
18	17	com12 32	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∪ 𝐵) = 𝐶 → (𝐶 ∖ 𝐴) = 𝐵))
19	18	adantl 485	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐴 ∪ 𝐵) = 𝐶 → (𝐶 ∖ 𝐴) = 𝐵))
20		simpl 486	. . . . . 6 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → 𝐴 ⊆ 𝐶)
21		difssd 4092	. . . . . . . 8 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → (𝐶 ∖ 𝐴) ⊆ 𝐶)
22		sseq1 3963	. . . . . . . 8 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → ((𝐶 ∖ 𝐴) ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
23	21, 22	mpbid 234	. . . . . . 7 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → 𝐵 ⊆ 𝐶)
24	23	adantl 485	. . . . . 6 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → 𝐵 ⊆ 𝐶)
25	20, 24	unssd 4146	. . . . 5 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → (𝐴 ∪ 𝐵) ⊆ 𝐶)
26		eqimss 3996	. . . . . . 7 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → (𝐶 ∖ 𝐴) ⊆ 𝐵)
27		ssundif 4443	. . . . . . 7 ⊢ (𝐶 ⊆ (𝐴 ∪ 𝐵) ↔ (𝐶 ∖ 𝐴) ⊆ 𝐵)
28	26, 27	sylibr 236	. . . . . 6 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → 𝐶 ⊆ (𝐴 ∪ 𝐵))
29	28	adantl 485	. . . . 5 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → 𝐶 ⊆ (𝐴 ∪ 𝐵))
30	25, 29	eqssd 3955	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → (𝐴 ∪ 𝐵) = 𝐶)
31	30	ex 416	. . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝐶 ∖ 𝐴) = 𝐵 → (𝐴 ∪ 𝐵) = 𝐶))
32	31	adantr 484	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐶 ∖ 𝐴) = 𝐵 → (𝐴 ∪ 𝐵) = 𝐶))
33	19, 32	impbid 214	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐴 ∪ 𝐵) = 𝐶 ↔ (𝐶 ∖ 𝐴) = 𝐵))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399   = wceq 1562   ∖ cdif 3903   ∪ cun 3904   ∩ cin 3905   ⊆ wss 3906  ∅c0 4287

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-ext 2736

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1565  df-fal 1575  df-ex 1802  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-ral 3079  df-rab 3417  df-v 3458  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4288

This theorem is referenced by:  fzdifsuc  13591  hashbclem  14467  lecldbas  23281  conndisj  23478  ptuncnv  23869  ptunhmeo  23870  cldsubg  24173  icopnfcld  24829  iocmnfcld  24830  voliunlem1  25614  icombl  25628  ioombl  25629  uniioombllem4  25650  ismbf3d  25718  lhop  26080  symgcom  33265  f1resfz0f1d  35468  subfacp1lem3  35537  subfacp1lem5  35539  pconnconn  35586  cvmscld  35628