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Theorem sscon 4115
Description: Contraposition law for subsets. Exercise 15 of [TakeutiZaring] p. 22. (Contributed by NM, 22-Mar-1998.)
Assertion
Ref Expression
sscon (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))

Proof of Theorem sscon
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 ssel 3961 . . . . 5 (𝐴𝐵 → (𝑥𝐴𝑥𝐵))
21con3d 155 . . . 4 (𝐴𝐵 → (¬ 𝑥𝐵 → ¬ 𝑥𝐴))
32anim2d 613 . . 3 (𝐴𝐵 → ((𝑥𝐶 ∧ ¬ 𝑥𝐵) → (𝑥𝐶 ∧ ¬ 𝑥𝐴)))
4 eldif 3946 . . 3 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
5 eldif 3946 . . 3 (𝑥 ∈ (𝐶𝐴) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐴))
63, 4, 53imtr4g 298 . 2 (𝐴𝐵 → (𝑥 ∈ (𝐶𝐵) → 𝑥 ∈ (𝐶𝐴)))
76ssrdv 3973 1 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 398  wcel 2110  cdif 3933  wss 3936
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 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2156  ax-12 2172  ax-ext 2793
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-v 3497  df-dif 3939  df-in 3943  df-ss 3952
This theorem is referenced by:  sscond  4118  complss  4123  sorpsscmpl  7454  sbthlem1  8621  sbthlem2  8622  cantnfp1lem1  9135  cantnfp1lem3  9137  isf34lem7  9795  isf34lem6  9796  setsres  16519  mplsubglem  20208  cctop  21608  clsval2  21652  ntrss  21657  hauscmplem  22008  ptbasin  22179  cfinfil  22495  csdfil  22496  uniioombllem5  24182  kur14lem6  32453  bj-2upln1upl  34331  dvasin  34972  sscon34b  40362  clsk3nimkb  40383  fourierdlem62  42446  caragendifcl  42789
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