Theorem sscon34b 4199

Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss 4052. (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 4044	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))
2		sscon 4044	. . 3 ⊢ ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → (𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵)))
3		dfss4 4163	. . . . . 6 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
4	3	biimpi 219	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
5	4	adantr 484	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
6		dfss4 4163	. . . . . 6 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
7	6	biimpi 219	. . . . 5 ⊢ (𝐵 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
8	7	adantl 485	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
9	5, 8	sseq12d 3925	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵)) ↔ 𝐴 ⊆ 𝐵))
10	2, 9	syl5ib 247	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → 𝐴 ⊆ 𝐵))
11	1, 10	impbid2 229	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399   = wceq 1538   ∖ cdif 3855   ⊆ wss 3858

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2729

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2736  df-cleq 2750  df-clel 2830  df-rab 3079  df-v 3411  df-dif 3861  df-in 3865  df-ss 3875

This theorem is referenced by:  rcompleq  4200  ntrclsss  41161  ntrclsiso  41165  ntrclsk2  41166  ntrclsk3  41168