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Theorem sscon34b 4254
Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss 4106. (Contributed by RP, 3-Jun-2021.)
Assertion
Ref Expression
sscon34b ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))

Proof of Theorem sscon34b
StepHypRef Expression
1 sscon 4098 . 2 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
2 sscon 4098 . . 3 ((𝐶𝐵) ⊆ (𝐶𝐴) → (𝐶 ∖ (𝐶𝐴)) ⊆ (𝐶 ∖ (𝐶𝐵)))
3 dfss4 4218 . . . . . 6 (𝐴𝐶 ↔ (𝐶 ∖ (𝐶𝐴)) = 𝐴)
43biimpi 215 . . . . 5 (𝐴𝐶 → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
54adantr 481 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
6 dfss4 4218 . . . . . 6 (𝐵𝐶 ↔ (𝐶 ∖ (𝐶𝐵)) = 𝐵)
76biimpi 215 . . . . 5 (𝐵𝐶 → (𝐶 ∖ (𝐶𝐵)) = 𝐵)
87adantl 482 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐵)) = 𝐵)
95, 8sseq12d 3977 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐶 ∖ (𝐶𝐴)) ⊆ (𝐶 ∖ (𝐶𝐵)) ↔ 𝐴𝐵))
102, 9imbitrid 243 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐶𝐵) ⊆ (𝐶𝐴) → 𝐴𝐵))
111, 10impbid2 225 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 396   = wceq 1541  cdif 3907  wss 3910
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 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2707
This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1544  df-ex 1782  df-sb 2068  df-clab 2714  df-cleq 2728  df-clel 2814  df-rab 3408  df-v 3447  df-dif 3913  df-in 3917  df-ss 3927
This theorem is referenced by:  rcompleq  4255  ntrclsss  42325  ntrclsiso  42329  ntrclsk2  42330  ntrclsk3  42332
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