Theorem eqdif 33163

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 3923	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
2		ssdif0 4313	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∖ 𝐵) = ∅)
3		ssdif0 4313	. . 3 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∖ 𝐴) = ∅)
4	2, 3	anbi12i 640	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ ((𝐴 ∖ 𝐵) = ∅ ∧ (𝐵 ∖ 𝐴) = ∅))
5	1, 4	sylbbr 239	1 ⊢ (((𝐴 ∖ 𝐵) = ∅ ∧ (𝐵 ∖ 𝐴) = ∅) → 𝐴 = 𝐵)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1569   ⊆ wss 3895   ∖ cdif 4045  ∅c0 4278

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-ss 3896  df-dif 4047  df-nul 4279

This theorem is used by:  pmtrcnelor  33699