MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  sscon34b Structured version   Visualization version   GIF version

Theorem sscon34b 4310
Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss 4161. (Contributed by RP, 3-Jun-2021.)
Assertion
Ref Expression
sscon34b ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))

Proof of Theorem sscon34b
StepHypRef Expression
1 sscon 4153 . 2 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
2 sscon 4153 . . 3 ((𝐶𝐵) ⊆ (𝐶𝐴) → (𝐶 ∖ (𝐶𝐴)) ⊆ (𝐶 ∖ (𝐶𝐵)))
3 dfss4 4275 . . . . . 6 (𝐴𝐶 ↔ (𝐶 ∖ (𝐶𝐴)) = 𝐴)
43biimpi 216 . . . . 5 (𝐴𝐶 → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
54adantr 480 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
6 dfss4 4275 . . . . . 6 (𝐵𝐶 ↔ (𝐶 ∖ (𝐶𝐵)) = 𝐵)
76biimpi 216 . . . . 5 (𝐵𝐶 → (𝐶 ∖ (𝐶𝐵)) = 𝐵)
87adantl 481 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐶 ∖ (𝐶𝐵)) = 𝐵)
95, 8sseq12d 4029 . . 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 1537  cdif 3960  wss 3963
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-dif 3966  df-in 3970  df-ss 3980
This theorem is referenced by:  rcompleq  4311  ntrclsss  44053  ntrclsiso  44057  ntrclsk2  44058  ntrclsk3  44060
  Copyright terms: Public domain W3C validator