Theorem eqdif 32395

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 3992	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
2		ssdif0 4363	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∖ 𝐵) = ∅)
3		ssdif0 4363	. . 3 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∖ 𝐴) = ∅)
4	2, 3	anbi12i 626	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((𝐴 ∖ 𝐵) = ∅ ∧ (𝐵 ∖ 𝐴) = ∅))
5	1, 4	sylbbr 235	1 ⊢ (((𝐴 ∖ 𝐵) = ∅ ∧ (𝐵 ∖ 𝐴) = ∅) → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ∧ wa 394   = wceq 1533   ∖ cdif 3941   ⊆ wss 3944  ∅c0 4322

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2696

This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1536  df-fal 1546  df-ex 1774  df-sb 2060  df-clab 2703  df-cleq 2717  df-clel 2802  df-v 3463  df-dif 3947  df-ss 3961  df-nul 4323

This theorem is referenced by:  pmtrcnelor  32904