Theorem disjeq0 4415

Description: Two disjoint sets are equal iff both are empty. (Contributed by AV, 19-Jun-2022.)

Assertion

Ref	Expression
disjeq0	⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 = 𝐵 ↔ (𝐴 = ∅ ∧ 𝐵 = ∅)))

Proof of Theorem disjeq0

Step	Hyp	Ref	Expression
1		ineq1 4165	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐵))
2		inidm 4178	. . . . . 6 ⊢ (𝐵 ∩ 𝐵) = 𝐵
3	1, 2	eqtrdi 2813	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = 𝐵)
4	3	eqeq1d 2764	. . . 4 ⊢ (𝐴 = 𝐵 → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐵 = ∅))
5		eqtr 2782	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = ∅) → 𝐴 = ∅)
6		simpr 489	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = ∅) → 𝐵 = ∅)
7	5, 6	jca 520	. . . . 5 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = ∅) → (𝐴 = ∅ ∧ 𝐵 = ∅))
8	7	ex 417	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐵 = ∅ → (𝐴 = ∅ ∧ 𝐵 = ∅)))
9	4, 8	sylbid 243	. . 3 ⊢ (𝐴 = 𝐵 → ((𝐴 ∩ 𝐵) = ∅ → (𝐴 = ∅ ∧ 𝐵 = ∅)))
10	9	com12 33	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 = 𝐵 → (𝐴 = ∅ ∧ 𝐵 = ∅)))
11		eqtr3 2784	. 2 ⊢ ((𝐴 = ∅ ∧ 𝐵 = ∅) → 𝐴 = 𝐵)
12	10, 11	impbid1 228	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 = 𝐵 ↔ (𝐴 = ∅ ∧ 𝐵 = ∅)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1569   ∩ cin 3903  ∅c0 4285

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3416  df-v 3456  df-in 3911

This theorem is used by:  epnsym  9576