Theorem eqdif 30866

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 3936	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
2		ssdif0 4297	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∖ 𝐵) = ∅)
3		ssdif0 4297	. . 3 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∖ 𝐴) = ∅)
4	2, 3	anbi12i 627	. 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 396   = wceq 1539   ∖ cdif 3884   ⊆ wss 3887  ∅c0 4256

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-dif 3890  df-in 3894  df-ss 3904  df-nul 4257

This theorem is referenced by:  pmtrcnelor  31360