Theorem ssdifim 4196

Description: Implication of a class difference with a subclass. (Contributed by AV, 3-Jan-2022.)

Assertion

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

Proof of Theorem ssdifim

Step	Hyp	Ref	Expression
1		dfss4 4192	. . 3 ⊢ (𝐴 ⊆ 𝑉 ↔ (𝑉 ∖ (𝑉 ∖ 𝐴)) = 𝐴)
2		eqcom 2745	. . 3 ⊢ ((𝑉 ∖ (𝑉 ∖ 𝐴)) = 𝐴 ↔ 𝐴 = (𝑉 ∖ (𝑉 ∖ 𝐴)))
3	1, 2	sylbb 218	. 2 ⊢ (𝐴 ⊆ 𝑉 → 𝐴 = (𝑉 ∖ (𝑉 ∖ 𝐴)))
4		difeq2 4051	. . 3 ⊢ (𝐵 = (𝑉 ∖ 𝐴) → (𝑉 ∖ 𝐵) = (𝑉 ∖ (𝑉 ∖ 𝐴)))
5	4	eqcomd 2744	. 2 ⊢ (𝐵 = (𝑉 ∖ 𝐴) → (𝑉 ∖ (𝑉 ∖ 𝐴)) = (𝑉 ∖ 𝐵))
6	3, 5	sylan9eq 2798	1 ⊢ ((𝐴 ⊆ 𝑉 ∧ 𝐵 = (𝑉 ∖ 𝐴)) → 𝐴 = (𝑉 ∖ 𝐵))

Syntax hints:   → wi 4   ∧ wa 396   = wceq 1539   ∖ cdif 3884   ⊆ wss 3887

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-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-rab 3073  df-v 3434  df-dif 3890  df-in 3894  df-ss 3904

This theorem is referenced by:  ssdifsym  4197  frgrwopregbsn  28681