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Theorem sscon34b 4232
Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss 4081. (Contributed by RP, 3-Jun-2021.)
Assertion
Ref Expression
sscon34b ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))

Proof of Theorem sscon34b
StepHypRef Expression
1 sscon 4073 . 2 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
2 sscon 4073 . . 3 ((𝐶𝐵) ⊆ (𝐶𝐴) → (𝐶 ∖ (𝐶𝐴)) ⊆ (𝐶 ∖ (𝐶𝐵)))
3 dfss4 4197 . . . . 5 (𝐴𝐶 ↔ (𝐶 ∖ (𝐶𝐴)) = 𝐴)
43birani 504 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
5 dfss4 4197 . . . . 5 (𝐵𝐶 ↔ (𝐶 ∖ (𝐶𝐵)) = 𝐵)
65bilani 505 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐵)) = 𝐵)
74, 6sseq12d 3948 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐶 ∖ (𝐶𝐴)) ⊆ (𝐶 ∖ (𝐶𝐵)) ↔ 𝐴𝐵))
82, 7imbitrid 245 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐶𝐵) ⊆ (𝐶𝐴) → 𝐴𝐵))
91, 8impbid2 227 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207  wa 396   = wceq 1547  cdif 3880  wss 3883
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711
This theorem depends on definitions:  df-bi 208  df-an 397  df-3an 1094  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-rab 3392  df-v 3433  df-dif 3886  df-in 3890  df-ss 3900
This theorem is referenced by:  rcompleq  4233  ntrclsss  44507  ntrclsiso  44511  ntrclsk2  44512  ntrclsk3  44514
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