Theorem unineq 4204

Description: Infer equality from equalities of union and intersection. Exercise 20 of [Enderton] p. 32 and its converse. (Contributed by NM, 10-Aug-2004.)

Assertion

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

Proof of Theorem unineq

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq2 2878	. . . . . . 7 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ 𝑥 ∈ (𝐵 ∩ 𝐶)))
2		elin 3897	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶))
3		elin 3897	. . . . . . 7 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
4	1, 2, 3	3bitr3g 316	. . . . . 6 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
5		iba 531	. . . . . . 7 ⊢ (𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
6		iba 531	. . . . . . 7 ⊢ (𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
7	5, 6	bibi12d 349	. . . . . 6 ⊢ (𝑥 ∈ 𝐶 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))))
8	4, 7	syl5ibr 249	. . . . 5 ⊢ (𝑥 ∈ 𝐶 → ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
9	8	adantld 494	. . . 4 ⊢ (𝑥 ∈ 𝐶 → (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
10		uncom 4080	. . . . . . . . 9 ⊢ (𝐴 ∪ 𝐶) = (𝐶 ∪ 𝐴)
11		uncom 4080	. . . . . . . . 9 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵)
12	10, 11	eqeq12i 2813	. . . . . . . 8 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ↔ (𝐶 ∪ 𝐴) = (𝐶 ∪ 𝐵))
13		eleq2 2878	. . . . . . . 8 ⊢ ((𝐶 ∪ 𝐴) = (𝐶 ∪ 𝐵) → (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ 𝑥 ∈ (𝐶 ∪ 𝐵)))
14	12, 13	sylbi 220	. . . . . . 7 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ 𝑥 ∈ (𝐶 ∪ 𝐵)))
15		elun 4076	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴))
16		elun 4076	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∪ 𝐵) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵))
17	14, 15, 16	3bitr3g 316	. . . . . 6 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → ((𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵)))
18		biorf 934	. . . . . . 7 ⊢ (¬ 𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴)))
19		biorf 934	. . . . . . 7 ⊢ (¬ 𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵)))
20	18, 19	bibi12d 349	. . . . . 6 ⊢ (¬ 𝑥 ∈ 𝐶 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵))))
21	17, 20	syl5ibr 249	. . . . 5 ⊢ (¬ 𝑥 ∈ 𝐶 → ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
22	21	adantrd 495	. . . 4 ⊢ (¬ 𝑥 ∈ 𝐶 → (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
23	9, 22	pm2.61i 185	. . 3 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
24	23	eqrdv 2796	. 2 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → 𝐴 = 𝐵)
25		uneq1 4083	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶))
26		ineq1 4131	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶))
27	25, 26	jca 515	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
28	24, 27	impbii 212	1 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) ↔ 𝐴 = 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399   ∨ wo 844   = wceq 1538   ∈ wcel 2111   ∪ cun 3879   ∩ cin 3880

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 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-rab 3115  df-v 3443  df-un 3886  df-in 3888

This theorem is referenced by: (None)