Theorem unineq 1680

Description: Deduce equality from equalities of union and intersection. Exercise 20 of [Enderton] p. 32 and its converse.

Assertion

Ref	Expression
unineq	⊢ (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) ↔ A = B)

Proof of Theorem unineq

Step	Hyp	Ref	Expression
1		iba 486	. . . . . . 7 ⊢ (x ∈ C → (x ∈ A ↔ (x ∈ A ∧ x ∈ C)))
2		iba 486	. . . . . . 7 ⊢ (x ∈ C → (x ∈ B ↔ (x ∈ B ∧ x ∈ C)))
3	1, 2	bibi12d 477	. . . . . 6 ⊢ (x ∈ C → ((x ∈ A ↔ x ∈ B) ↔ ((x ∈ A ∧ x ∈ C) ↔ (x ∈ B ∧ x ∈ C))))
4		eleq2 1150	. . . . . . 7 ⊢ ((A ∩ C) = (B ∩ C) → (x ∈ (A ∩ C) ↔ x ∈ (B ∩ C)))
5		elin 1635	. . . . . . 7 ⊢ (x ∈ (A ∩ C) ↔ (x ∈ A ∧ x ∈ C))
6		elin 1635	. . . . . . 7 ⊢ (x ∈ (B ∩ C) ↔ (x ∈ B ∧ x ∈ C))
7	4, 5, 6	3bitr3g 427	. . . . . 6 ⊢ ((A ∩ C) = (B ∩ C) → ((x ∈ A ∧ x ∈ C) ↔ (x ∈ B ∧ x ∈ C)))
8	3, 7	syl5bir 184	. . . . 5 ⊢ (x ∈ C → ((A ∩ C) = (B ∩ C) → (x ∈ A ↔ x ∈ B)))
9	8	adantld 307	. . . 4 ⊢ (x ∈ C → (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) → (x ∈ A ↔ x ∈ B)))
10		biorf 551	. . . . . . 7 ⊢ (¬ x ∈ C → (x ∈ A ↔ (x ∈ C ∨ x ∈ A)))
11		biorf 551	. . . . . . 7 ⊢ (¬ x ∈ C → (x ∈ B ↔ (x ∈ C ∨ x ∈ B)))
12	10, 11	bibi12d 477	. . . . . 6 ⊢ (¬ x ∈ C → ((x ∈ A ↔ x ∈ B) ↔ ((x ∈ C ∨ x ∈ A) ↔ (x ∈ C ∨ x ∈ B))))
13		uncom 1604	. . . . . . . . 9 ⊢ (A ∪ C) = (C ∪ A)
14		uncom 1604	. . . . . . . . 9 ⊢ (B ∪ C) = (C ∪ B)
15	13, 14	cleq12i 1114	. . . . . . . 8 ⊢ ((A ∪ C) = (B ∪ C) ↔ (C ∪ A) = (C ∪ B))
16		eleq2 1150	. . . . . . . 8 ⊢ ((C ∪ A) = (C ∪ B) → (x ∈ (C ∪ A) ↔ x ∈ (C ∪ B)))
17	15, 16	sylbi 174	. . . . . . 7 ⊢ ((A ∪ C) = (B ∪ C) → (x ∈ (C ∪ A) ↔ x ∈ (C ∪ B)))
18		elun 1601	. . . . . . 7 ⊢ (x ∈ (C ∪ A) ↔ (x ∈ C ∨ x ∈ A))
19		elun 1601	. . . . . . 7 ⊢ (x ∈ (C ∪ B) ↔ (x ∈ C ∨ x ∈ B))
20	17, 18, 19	3bitr3g 427	. . . . . 6 ⊢ ((A ∪ C) = (B ∪ C) → ((x ∈ C ∨ x ∈ A) ↔ (x ∈ C ∨ x ∈ B)))
21	12, 20	syl5bir 184	. . . . 5 ⊢ (¬ x ∈ C → ((A ∪ C) = (B ∪ C) → (x ∈ A ↔ x ∈ B)))
22	21	adantrd 308	. . . 4 ⊢ (¬ x ∈ C → (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) → (x ∈ A ↔ x ∈ B)))
23	9, 22	pm2.61i 110	. . 3 ⊢ (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) → (x ∈ A ↔ x ∈ B))
24	23	cleqrd 1100	. 2 ⊢ (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) → A = B)
25		uneq1 1605	. . 3 ⊢ (A = B → (A ∪ C) = (B ∪ C))
26		ineq1 1638	. . 3 ⊢ (A = B → (A ∩ C) = (B ∩ C))
27	25, 26	jca 236	. 2 ⊢ (A = B → ((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)))
28	24, 27	impbi 139	1 ⊢ (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) ↔ A = B)

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   ∨ wo 195   ∧ wa 196   = wceq 1091   ∈ wcel 1092   ∪ cun 1485   ∩ cin 1486

This theorem is referenced by:  mapdom2 3389

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490  df-in 1491

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