Theorem ssdifsym 4223

Description: Symmetric class differences for subclasses. (Contributed by AV, 3-Jan-2022.)

Assertion

Ref	Expression
ssdifsym	⊢ ((𝐴 ⊆ 𝑉 ∧ 𝐵 ⊆ 𝑉) → (𝐵 = (𝑉 ∖ 𝐴) ↔ 𝐴 = (𝑉 ∖ 𝐵)))

Proof of Theorem ssdifsym

Step	Hyp	Ref	Expression
1		ssdifim 4222	. . 3 ⊢ ((𝐴 ⊆ 𝑉 ∧ 𝐵 = (𝑉 ∖ 𝐴)) → 𝐴 = (𝑉 ∖ 𝐵))
2	1	ex 418	. 2 ⊢ (𝐴 ⊆ 𝑉 → (𝐵 = (𝑉 ∖ 𝐴) → 𝐴 = (𝑉 ∖ 𝐵)))
3		ssdifim 4222	. . 3 ⊢ ((𝐵 ⊆ 𝑉 ∧ 𝐴 = (𝑉 ∖ 𝐵)) → 𝐵 = (𝑉 ∖ 𝐴))
4	3	ex 418	. 2 ⊢ (𝐵 ⊆ 𝑉 → (𝐴 = (𝑉 ∖ 𝐵) → 𝐵 = (𝑉 ∖ 𝐴)))
5	2, 4	anbiim 653	1 ⊢ ((𝐴 ⊆ 𝑉 ∧ 𝐵 ⊆ 𝑉) → (𝐵 = (𝑉 ∖ 𝐴) ↔ 𝐴 = (𝑉 ∖ 𝐵)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∖ cdif 3899   ⊆ wss 3902

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3455  df-dif 3905  df-in 3909  df-ss 3919

This theorem is used by:  zarcls  34371