Theorem unineq 4268

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 2824	. . . . . . 7 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ 𝑥 ∈ (𝐵 ∩ 𝐶)))
2		elin 3947	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶))
3		elin 3947	. . . . . . 7 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
4	1, 2, 3	3bitr3g 313	. . . . . 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	imbitrrid 246	. . . . 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 4138	. . . . . . . . 9 ⊢ (𝐴 ∪ 𝐶) = (𝐶 ∪ 𝐴)
11		uncom 4138	. . . . . . . . 9 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵)
12	10, 11	eqeq12i 2754	. . . . . . . 8 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ↔ (𝐶 ∪ 𝐴) = (𝐶 ∪ 𝐵))
13		eleq2 2824	. . . . . . . 8 ⊢ ((𝐶 ∪ 𝐴) = (𝐶 ∪ 𝐵) → (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ 𝑥 ∈ (𝐶 ∪ 𝐵)))
14	12, 13	sylbi 217	. . . . . . 7 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ 𝑥 ∈ (𝐶 ∪ 𝐵)))
15		elun 4133	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴))
16		elun 4133	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∪ 𝐵) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵))
17	14, 15, 16	3bitr3g 313	. . . . . 6 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → ((𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵)))
18		biorf 936	. . . . . . 7 ⊢ (¬ 𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴)))
19		biorf 936	. . . . . . 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	imbitrrid 246	. . . . 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 2734	. 2 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → 𝐴 = 𝐵)
25		uneq1 4141	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶))
26		ineq1 4193	. . 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 209	1 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) ↔ 𝐴 = 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395   ∨ wo 847   = wceq 1540   ∈ wcel 2109   ∪ cun 3929   ∩ cin 3930

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810  df-rab 3421  df-v 3466  df-un 3936  df-in 3938

This theorem is referenced by: (None)