Theorem unineq 3877

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 2690	. . . . . . 7 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ 𝑥 ∈ (𝐵 ∩ 𝐶)))
2		elin 3796	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶))
3		elin 3796	. . . . . . 7 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
4	1, 2, 3	3bitr3g 302	. . . . . 6 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
5		iba 524	. . . . . . 7 ⊢ (𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
6		iba 524	. . . . . . 7 ⊢ (𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
7	5, 6	bibi12d 335	. . . . . 6 ⊢ (𝑥 ∈ 𝐶 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))))
8	4, 7	syl5ibr 236	. . . . 5 ⊢ (𝑥 ∈ 𝐶 → ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
9	8	adantld 483	. . . 4 ⊢ (𝑥 ∈ 𝐶 → (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
10		uncom 3757	. . . . . . . . 9 ⊢ (𝐴 ∪ 𝐶) = (𝐶 ∪ 𝐴)
11		uncom 3757	. . . . . . . . 9 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵)
12	10, 11	eqeq12i 2636	. . . . . . . 8 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ↔ (𝐶 ∪ 𝐴) = (𝐶 ∪ 𝐵))
13		eleq2 2690	. . . . . . . 8 ⊢ ((𝐶 ∪ 𝐴) = (𝐶 ∪ 𝐵) → (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ 𝑥 ∈ (𝐶 ∪ 𝐵)))
14	12, 13	sylbi 207	. . . . . . 7 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ 𝑥 ∈ (𝐶 ∪ 𝐵)))
15		elun 3753	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∪ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴))
16		elun 3753	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∪ 𝐵) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵))
17	14, 15, 16	3bitr3g 302	. . . . . 6 ⊢ ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → ((𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵)))
18		biorf 420	. . . . . . 7 ⊢ (¬ 𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴)))
19		biorf 420	. . . . . . 7 ⊢ (¬ 𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵)))
20	18, 19	bibi12d 335	. . . . . 6 ⊢ (¬ 𝑥 ∈ 𝐶 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∨ 𝑥 ∈ 𝐵))))
21	17, 20	syl5ibr 236	. . . . 5 ⊢ (¬ 𝑥 ∈ 𝐶 → ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
22	21	adantrd 484	. . . 4 ⊢ (¬ 𝑥 ∈ 𝐶 → (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
23	9, 22	pm2.61i 176	. . 3 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
24	23	eqrdv 2620	. 2 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → 𝐴 = 𝐵)
25		uneq1 3760	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶))
26		ineq1 3807	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶))
27	25, 26	jca 554	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
28	24, 27	impbii 199	1 ⊢ (((𝐴 ∪ 𝐶) = (𝐵 ∪ 𝐶) ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) ↔ 𝐴 = 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 196   ∨ wo 383   ∧ wa 384   = wceq 1483   ∈ wcel 1990   ∪ cun 3572   ∩ cin 3573

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-un 3579  df-in 3581

This theorem is referenced by: (None)