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Theorem sscon 3132
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 3017 . . . . 5 (𝐴𝐵 → (𝑥𝐴𝑥𝐵))
21con3d 596 . . . 4 (𝐴𝐵 → (¬ 𝑥𝐵 → ¬ 𝑥𝐴))
32anim2d 330 . . 3 (𝐴𝐵 → ((𝑥𝐶 ∧ ¬ 𝑥𝐵) → (𝑥𝐶 ∧ ¬ 𝑥𝐴)))
4 eldif 3006 . . 3 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
5 eldif 3006 . . 3 (𝑥 ∈ (𝐶𝐴) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐴))
63, 4, 53imtr4g 203 . 2 (𝐴𝐵 → (𝑥 ∈ (𝐶𝐵) → 𝑥 ∈ (𝐶𝐴)))
76ssrdv 3029 1 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 102  wcel 1438  cdif 2994  wss 2997
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 579  ax-in2 580  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-tru 1292  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-v 2621  df-dif 2999  df-in 3003  df-ss 3010
This theorem is referenced by:  sscond  3135  sbthlem1  6645  sbthlem2  6646
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