Theorem sscon34b 38317

Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss 3751. (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 3744	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))
2		sscon 3744	. . 3 ⊢ ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → (𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵)))
3		dfss4 3858	. . . . . 6 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
4	3	biimpi 206	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
5	4	adantr 481	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
6		dfss4 3858	. . . . . 6 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
7	6	biimpi 206	. . . . 5 ⊢ (𝐵 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
8	7	adantl 482	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐶 ∖ (𝐶 ∖ 𝐵)) = 𝐵)
9	5, 8	sseq12d 3634	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ⊆ (𝐶 ∖ (𝐶 ∖ 𝐵)) ↔ 𝐴 ⊆ 𝐵))
10	2, 9	syl5ib 234	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴) → 𝐴 ⊆ 𝐵))
11	1, 10	impbid2 216	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐴 ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴)))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384   = wceq 1483   ∖ cdif 3571   ⊆ wss 3574

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-dif 3577  df-in 3581  df-ss 3588

This theorem is referenced by:  rcompleq  38318  ntrclsss  38361  ntrclsiso  38365  ntrclsk2  38366  ntrclsk3  38368