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Theorem sscon 4113
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 3959 . . . . 5 (𝐴𝐵 → (𝑥𝐴𝑥𝐵))
21con3d 155 . . . 4 (𝐴𝐵 → (¬ 𝑥𝐵 → ¬ 𝑥𝐴))
32anim2d 613 . . 3 (𝐴𝐵 → ((𝑥𝐶 ∧ ¬ 𝑥𝐵) → (𝑥𝐶 ∧ ¬ 𝑥𝐴)))
4 eldif 3944 . . 3 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
5 eldif 3944 . . 3 (𝑥 ∈ (𝐶𝐴) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐴))
63, 4, 53imtr4g 298 . 2 (𝐴𝐵 → (𝑥 ∈ (𝐶𝐵) → 𝑥 ∈ (𝐶𝐴)))
76ssrdv 3971 1 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 398  wcel 2108  cdif 3931  wss 3934
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1905  ax-6 1964  ax-7 2009  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2154  ax-12 2170  ax-ext 2791
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1534  df-ex 1775  df-nf 1779  df-sb 2064  df-clab 2798  df-cleq 2812  df-clel 2891  df-nfc 2961  df-v 3495  df-dif 3937  df-in 3941  df-ss 3950
This theorem is referenced by:  sscond  4116  complss  4121  sorpsscmpl  7452  sbthlem1  8619  sbthlem2  8620  cantnfp1lem1  9133  cantnfp1lem3  9135  isf34lem7  9793  isf34lem6  9794  setsres  16517  mplsubglem  20206  cctop  21606  clsval2  21650  ntrss  21655  hauscmplem  22006  ptbasin  22177  cfinfil  22493  csdfil  22494  uniioombllem5  24180  kur14lem6  32451  bj-2upln1upl  34329  dvasin  34970  sscon34b  40359  clsk3nimkb  40380  fourierdlem62  42443  caragendifcl  42786
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