Theorem unineq 4208

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 2827	. . . . . . 7 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ 𝑥 ∈ (𝐵 ∩ 𝐶)))
2		elin 3899	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶))
3		elin 3899	. . . . . . 7 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
4	1, 2, 3	3bitr3g 312	. . . . . 6 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
5		iba 527	. . . . . . 7 ⊢ (𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
6		iba 527	. . . . . . 7 ⊢ (𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
7	5, 6	bibi12d 345	. . . . . 6 ⊢ (𝑥 ∈ 𝐶 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))))
8	4, 7	syl5ibr 245	. . . . 5 ⊢ (𝑥 ∈ 𝐶 → ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
9	8	adantld 490	. . . 4 ⊢ (𝑥 ∈ 𝐶 → (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
10		uncom 4083	. . . . . . . . 9 ⊢ (𝐴 ∪ 𝐶) = (𝐶 ∪ 𝐴)
11		uncom 4083	. . . . . . . . 9 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵)
12	10, 11	eqeq12i 2756	. . . . . . . 8 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ↔ (𝐶 ∪ 𝐴) = (𝐶 ∪ 𝐵))
13		eleq2 2827	. . . . . . . 8 ⊢ ((𝐶 ∪ 𝐴) = (𝐶 ∪ 𝐵) → (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ 𝑥 ∈ (𝐶 ∪ 𝐵)))
14	12, 13	sylbi 216	. . . . . . 7 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ 𝑥 ∈ (𝐶 ∪ 𝐵)))
15		elun 4079	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴))
16		elun 4079	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∪ 𝐵) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵))
17	14, 15, 16	3bitr3g 312	. . . . . 6 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → ((𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵)))
18		biorf 933	. . . . . . 7 ⊢ (¬ 𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴)))
19		biorf 933	. . . . . . 7 ⊢ (¬ 𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵)))
20	18, 19	bibi12d 345	. . . . . 6 ⊢ (¬ 𝑥 ∈ 𝐶 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵))))
21	17, 20	syl5ibr 245	. . . . 5 ⊢ (¬ 𝑥 ∈ 𝐶 → ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
22	21	adantrd 491	. . . 4 ⊢ (¬ 𝑥 ∈ 𝐶 → (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
23	9, 22	pm2.61i 182	. . 3 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
24	23	eqrdv 2736	. 2 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → 𝐴 = 𝐵)
25		uneq1 4086	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶))
26		ineq1 4136	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶))
27	25, 26	jca 511	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
28	24, 27	impbii 208	1 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) ↔ 𝐴 = 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 395   ∨ wo 843   = wceq 1539   ∈ wcel 2108   ∪ cun 3881   ∩ cin 3882

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-rab 3072  df-v 3424  df-un 3888  df-in 3890

This theorem is referenced by: (None)