Theorem uneqdifeq 4447

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 4112	. . . . 5 ⊢ (𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵)
2		eqtr 2757	. . . . . . 7 ⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → (𝐵 ∪ 𝐴) = 𝐶)
3	2	eqcomd 2743	. . . . . 6 ⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → 𝐶 = (𝐵 ∪ 𝐴))
4		difeq1 4073	. . . . . . 7 ⊢ (𝐶 = (𝐵 ∪ 𝐴) → (𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴))
5		difun2 4435	. . . . . . 7 ⊢ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)
6		eqtr 2757	. . . . . . . 8 ⊢ (((𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴) ∧ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)) → (𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴))
7		incom 4163	. . . . . . . . . . 11 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
8	7	eqeq1i 2742	. . . . . . . . . 10 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅)
9		disj3 4408	. . . . . . . . . 10 ⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ 𝐵 = (𝐵 ∖ 𝐴))
10	8, 9	bitri 275	. . . . . . . . 9 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐵 = (𝐵 ∖ 𝐴))
11		eqtr 2757	. . . . . . . . . . 11 ⊢ (((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) ∧ (𝐵 ∖ 𝐴) = 𝐵) → (𝐶 ∖ 𝐴) = 𝐵)
12	11	expcom 413	. . . . . . . . . 10 ⊢ ((𝐵 ∖ 𝐴) = 𝐵 → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵))
13	12	eqcoms 2745	. . . . . . . . 9 ⊢ (𝐵 = (𝐵 ∖ 𝐴) → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵))
14	10, 13	sylbi 217	. . . . . . . 8 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵))
15	6, 14	syl5com 31	. . . . . . 7 ⊢ (((𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴) ∧ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
16	4, 5, 15	sylancl 587	. . . . . 6 ⊢ (𝐶 = (𝐵 ∪ 𝐴) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
17	3, 16	syl 17	. . . . 5 ⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
18	1, 17	mpan 691	. . . 4 ⊢ ((𝐴 ∪ 𝐵) = 𝐶 → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
19	18	com12 32	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∪ 𝐵) = 𝐶 → (𝐶 ∖ 𝐴) = 𝐵))
20	19	adantl 481	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐴 ∪ 𝐵) = 𝐶 → (𝐶 ∖ 𝐴) = 𝐵))
21		simpl 482	. . . . . 6 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → 𝐴 ⊆ 𝐶)
22		difssd 4091	. . . . . . . 8 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → (𝐶 ∖ 𝐴) ⊆ 𝐶)
23		sseq1 3961	. . . . . . . 8 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → ((𝐶 ∖ 𝐴) ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
24	22, 23	mpbid 232	. . . . . . 7 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → 𝐵 ⊆ 𝐶)
25	24	adantl 481	. . . . . 6 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → 𝐵 ⊆ 𝐶)
26	21, 25	unssd 4146	. . . . 5 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → (𝐴 ∪ 𝐵) ⊆ 𝐶)
27		eqimss 3994	. . . . . . 7 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → (𝐶 ∖ 𝐴) ⊆ 𝐵)
28		ssundif 4442	. . . . . . 7 ⊢ (𝐶 ⊆ (𝐴 ∪ 𝐵) ↔ (𝐶 ∖ 𝐴) ⊆ 𝐵)
29	27, 28	sylibr 234	. . . . . 6 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → 𝐶 ⊆ (𝐴 ∪ 𝐵))
30	29	adantl 481	. . . . 5 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → 𝐶 ⊆ (𝐴 ∪ 𝐵))
31	26, 30	eqssd 3953	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → (𝐴 ∪ 𝐵) = 𝐶)
32	31	ex 412	. . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝐶 ∖ 𝐴) = 𝐵 → (𝐴 ∪ 𝐵) = 𝐶))
33	32	adantr 480	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐶 ∖ 𝐴) = 𝐵 → (𝐴 ∪ 𝐵) = 𝐶))
34	20, 33	impbid 212	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐴 ∪ 𝐵) = 𝐶 ↔ (𝐶 ∖ 𝐴) = 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1542   ∖ cdif 3900   ∪ cun 3901   ∩ cin 3902   ⊆ wss 3903  ∅c0 4287

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288

This theorem is referenced by:  fzdifsuc  13514  hashbclem  14389  lecldbas  23180  conndisj  23377  ptuncnv  23768  ptunhmeo  23769  cldsubg  24072  icopnfcld  24728  iocmnfcld  24729  voliunlem1  25524  icombl  25538  ioombl  25539  uniioombllem4  25560  ismbf3d  25628  lhop  25994  symgcom  33183  f1resfz0f1d  35336  subfacp1lem3  35404  subfacp1lem5  35406  pconnconn  35453  cvmscld  35495