Theorem eqdif 32494

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 3950	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
2		ssdif0 4316	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∖ 𝐵) = ∅)
3		ssdif0 4316	. . 3 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∖ 𝐴) = ∅)
4	2, 3	anbi12i 628	. 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 1541   ∖ cdif 3899   ⊆ wss 3902  ∅c0 4283

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-v 3438  df-dif 3905  df-ss 3919  df-nul 4284

This theorem is referenced by:  pmtrcnelor  33055