Theorem sscon34b 4270

Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss 4117. (Contributed by RP, 3-Jun-2021.)

Assertion

Ref	Expression
sscon34b	⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))

Proof of Theorem sscon34b

Step	Hyp	Ref	Expression
1		sscon 4109	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))
2		sscon 4109	. . 3 ⊢ ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → (𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵)))
3		dfss4 4235	. . . . . 6 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
4	3	biimpi 216	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
5	4	adantr 480	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
6		dfss4 4235	. . . . . 6 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
7	6	biimpi 216	. . . . 5 ⊢ (𝐵 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
8	7	adantl 481	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
9	5, 8	sseq12d 3983	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵)) ↔ 𝐴 ⊆ 𝐵))
10	2, 9	imbitrid 244	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → 𝐴 ⊆ 𝐵))
11	1, 10	impbid2 226	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1540   ∖ cdif 3914   ⊆ wss 3917

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-rab 3409  df-v 3452  df-dif 3920  df-in 3924  df-ss 3934

This theorem is referenced by:  rcompleq  4271  ntrclsss  44059  ntrclsiso  44063  ntrclsk2  44064  ntrclsk3  44066