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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 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))

Proof of Theorem sscon34b
StepHypRef Expression
1 sscon 4044 . 2 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
2 sscon 4044 . . 3 ((𝐶𝐵) ⊆ (𝐶𝐴) → (𝐶 ∖ (𝐶𝐴)) ⊆ (𝐶 ∖ (𝐶𝐵)))
3 dfss4 4163 . . . . . 6 (𝐴𝐶 ↔ (𝐶 ∖ (𝐶𝐴)) = 𝐴)
43biimpi 219 . . . . 5 (𝐴𝐶 → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
54adantr 484 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
6 dfss4 4163 . . . . . 6 (𝐵𝐶 ↔ (𝐶 ∖ (𝐶𝐵)) = 𝐵)
76biimpi 219 . . . . 5 (𝐵𝐶 → (𝐶 ∖ (𝐶𝐵)) = 𝐵)
87adantl 485 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐵)) = 𝐵)
95, 8sseq12d 3925 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐶 ∖ (𝐶𝐴)) ⊆ (𝐶 ∖ (𝐶𝐵)) ↔ 𝐴𝐵))
102, 9syl5ib 247 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐶𝐵) ⊆ (𝐶𝐴) → 𝐴𝐵))
111, 10impbid2 229 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
 Colors of variables: wff setvar class 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
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