Theorem sscon34b 40362

Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss 4122. (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 4114	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))
2		sscon 4114	. . 3 ⊢ ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → (𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵)))
3		dfss4 4234	. . . . . 6 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
4	3	biimpi 218	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
5	4	adantr 483	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
6		dfss4 4234	. . . . . 6 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
7	6	biimpi 218	. . . . 5 ⊢ (𝐵 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
8	7	adantl 484	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
9	5, 8	sseq12d 3999	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵)) ↔ 𝐴 ⊆ 𝐵))
10	2, 9	syl5ib 246	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → 𝐴 ⊆ 𝐵))
11	1, 10	impbid2 228	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398   = wceq 1533   ∖ cdif 3932   ⊆ wss 3935

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-rab 3147  df-v 3496  df-dif 3938  df-in 3942  df-ss 3951

This theorem is referenced by:  rcompleq  40363  ntrclsss  40406  ntrclsiso  40410  ntrclsk2  40411  ntrclsk3  40413