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Theorem sscon34b 4304
Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss 4151. (Contributed by RP, 3-Jun-2021.)
Assertion
Ref Expression
sscon34b ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))

Proof of Theorem sscon34b
StepHypRef Expression
1 sscon 4143 . 2 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
2 sscon 4143 . . 3 ((𝐶𝐵) ⊆ (𝐶𝐴) → (𝐶 ∖ (𝐶𝐴)) ⊆ (𝐶 ∖ (𝐶𝐵)))
3 dfss4 4269 . . . . . 6 (𝐴𝐶 ↔ (𝐶 ∖ (𝐶𝐴)) = 𝐴)
43biimpi 216 . . . . 5 (𝐴𝐶 → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
54adantr 480 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
6 dfss4 4269 . . . . . 6 (𝐵𝐶 ↔ (𝐶 ∖ (𝐶𝐵)) = 𝐵)
76biimpi 216 . . . . 5 (𝐵𝐶 → (𝐶 ∖ (𝐶𝐵)) = 𝐵)
87adantl 481 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐵)) = 𝐵)
95, 8sseq12d 4017 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐶 ∖ (𝐶𝐴)) ⊆ (𝐶 ∖ (𝐶𝐵)) ↔ 𝐴𝐵))
102, 9imbitrid 244 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐶𝐵) ⊆ (𝐶𝐴) → 𝐴𝐵))
111, 10impbid2 226 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1540  cdif 3948  wss 3951
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 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-rab 3437  df-v 3482  df-dif 3954  df-in 3958  df-ss 3968
This theorem is referenced by:  rcompleq  4305  ntrclsss  44076  ntrclsiso  44080  ntrclsk2  44081  ntrclsk3  44083
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