Theorem uneqdifeq 4451

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 4108	. . . . 5 ⊢ (𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵)
2		eqtr 2782	. . . . . . 7 ⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → (𝐵 ∪ 𝐴) = 𝐶)
3	2	eqcomd 2768	. . . . . 6 ⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → 𝐶 = (𝐵 ∪ 𝐴))
4		difeq1 4070	. . . . . . 7 ⊢ (𝐶 = (𝐵 ∪ 𝐴) → (𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴))
5		difun2 4440	. . . . . . 7 ⊢ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)
6		eqtr 2782	. . . . . . . 8 ⊢ (((𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴) ∧ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)) → (𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴))
7		ineqcom 4159	. . . . . . . . . 10 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅)
8		disj3 4410	. . . . . . . . . 10 ⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ 𝐵 = (𝐵 ∖ 𝐴))
9	7, 8	bitri 278	. . . . . . . . 9 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐵 = (𝐵 ∖ 𝐴))
10		eqtr 2782	. . . . . . . . . . 11 ⊢ (((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) ∧ (𝐵 ∖ 𝐴) = 𝐵) → (𝐶 ∖ 𝐴) = 𝐵)
11	10	expcom 419	. . . . . . . . . 10 ⊢ ((𝐵 ∖ 𝐴) = 𝐵 → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵))
12	11	eqcoms 2770	. . . . . . . . 9 ⊢ (𝐵 = (𝐵 ∖ 𝐴) → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵))
13	9, 12	sylbi 220	. . . . . . . 8 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵))
14	6, 13	syl5com 32	. . . . . . 7 ⊢ (((𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴) ∧ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
15	4, 5, 14	sylancl 598	. . . . . 6 ⊢ (𝐶 = (𝐵 ∪ 𝐴) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
16	3, 15	syl 18	. . . . 5 ⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
17	1, 16	mpan 703	. . . 4 ⊢ ((𝐴 ∪ 𝐵) = 𝐶 → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵))
18	17	com12 33	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∪ 𝐵) = 𝐶 → (𝐶 ∖ 𝐴) = 𝐵))
19	18	adantl 487	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐴 ∪ 𝐵) = 𝐶 → (𝐶 ∖ 𝐴) = 𝐵))
20		simpl 488	. . . . . 6 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → 𝐴 ⊆ 𝐶)
21		difssd 4087	. . . . . . . 8 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → (𝐶 ∖ 𝐴) ⊆ 𝐶)
22		sseq1 3959	. . . . . . . 8 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → ((𝐶 ∖ 𝐴) ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
23	21, 22	mpbid 235	. . . . . . 7 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → 𝐵 ⊆ 𝐶)
24	23	adantl 487	. . . . . 6 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → 𝐵 ⊆ 𝐶)
25	20, 24	unssd 4141	. . . . 5 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → (𝐴 ∪ 𝐵) ⊆ 𝐶)
26		eqimss 3992	. . . . . . 7 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → (𝐶 ∖ 𝐴) ⊆ 𝐵)
27		ssundif 4446	. . . . . . 7 ⊢ (𝐶 ⊆ (𝐴 ∪ 𝐵) ↔ (𝐶 ∖ 𝐴) ⊆ 𝐵)
28	26, 27	sylibr 237	. . . . . 6 ⊢ ((𝐶 ∖ 𝐴) = 𝐵 → 𝐶 ⊆ (𝐴 ∪ 𝐵))
29	28	adantl 487	. . . . 5 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → 𝐶 ⊆ (𝐴 ∪ 𝐵))
30	25, 29	eqssd 3951	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐶 ∖ 𝐴) = 𝐵) → (𝐴 ∪ 𝐵) = 𝐶)
31	30	ex 418	. . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝐶 ∖ 𝐴) = 𝐵 → (𝐴 ∪ 𝐵) = 𝐶))
32	31	adantr 486	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐶 ∖ 𝐴) = 𝐵 → (𝐴 ∪ 𝐵) = 𝐶))
33	19, 32	impbid 215	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐴 ∪ 𝐵) = 𝐶 ↔ (𝐶 ∖ 𝐴) = 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∖ cdif 3899   ∪ cun 3900   ∩ cin 3901   ⊆ wss 3902  ∅c0 4282

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283

This theorem is used by:  fzdifsuc  13641  f1resfz0f1d  13850  hashbclem  14519  lecldbas  23445  conndisj  23642  ptuncnv  24034  ptunhmeo  24035  cldsubg  24338  icopnfcld  24994  iocmnfcld  24995  voliunlem1  25779  icombl  25793  ioombl  25794  uniioombllem4  25815  ismbf3d  25883  lhop  26245  symgcom  33510  subfacp1lem3  35748  subfacp1lem5  35750  pconnconn  35797  cvmscld  35839