Theorem ssdifim 4193

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 4189	. . 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 4047	. . 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 2799	1 ⊢ ((𝐴 ⊆ 𝑉 ∧ 𝐵 = (𝑉 ∖ 𝐴)) → 𝐴 = (𝑉 ∖ 𝐵))

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1539   ∖ cdif 3880   ⊆ wss 3883

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-rab 3072  df-v 3424  df-dif 3886  df-in 3890  df-ss 3900

This theorem is referenced by:  ssdifsym  4194  frgrwopregbsn  28582