Theorem eqdif 32549

Description: If both set differences of two sets are empty, those sets are equal. (Contributed by Thierry Arnoux, 16-Nov-2023.)

Assertion

Ref	Expression
eqdif	⊢ (((𝐴 ∖ 𝐵) = ∅ ∧ (𝐵 ∖ 𝐴) = ∅) → 𝐴 = 𝐵)

Proof of Theorem eqdif

Step	Hyp	Ref	Expression
1		eqss 4024	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
2		ssdif0 4389	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∖ 𝐵) = ∅)
3		ssdif0 4389	. . 3 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∖ 𝐴) = ∅)
4	2, 3	anbi12i 627	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((𝐴 ∖ 𝐵) = ∅ ∧ (𝐵 ∖ 𝐴) = ∅))
5	1, 4	sylbbr 236	1 ⊢ (((𝐴 ∖ 𝐵) = ∅ ∧ (𝐵 ∖ 𝐴) = ∅) → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1537   ∖ cdif 3973   ⊆ wss 3976  ∅c0 4352

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-v 3490  df-dif 3979  df-ss 3993  df-nul 4353

This theorem is referenced by:  pmtrcnelor  33084