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Theorem sscon 3102
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 2964 . . . . 5 (𝐴𝐵 → (𝑥𝐴𝑥𝐵))
21con3d 569 . . . 4 (𝐴𝐵 → (¬ 𝑥𝐵 → ¬ 𝑥𝐴))
32anim2d 324 . . 3 (𝐴𝐵 → ((𝑥𝐶 ∧ ¬ 𝑥𝐵) → (𝑥𝐶 ∧ ¬ 𝑥𝐴)))
4 eldif 2952 . . 3 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
5 eldif 2952 . . 3 (𝑥 ∈ (𝐶𝐴) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐴))
63, 4, 53imtr4g 198 . 2 (𝐴𝐵 → (𝑥 ∈ (𝐶𝐵) → 𝑥 ∈ (𝐶𝐴)))
76ssrdv 2976 1 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 101  wcel 1407  cdif 2939  wss 2942
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-in1 552  ax-in2 553  ax-io 638  ax-5 1350  ax-7 1351  ax-gen 1352  ax-ie1 1396  ax-ie2 1397  ax-8 1409  ax-10 1410  ax-11 1411  ax-i12 1412  ax-bndl 1413  ax-4 1414  ax-17 1433  ax-i9 1437  ax-ial 1441  ax-i5r 1442  ax-ext 2036
This theorem depends on definitions:  df-bi 114  df-tru 1260  df-nf 1364  df-sb 1660  df-clab 2041  df-cleq 2047  df-clel 2050  df-nfc 2181  df-v 2574  df-dif 2945  df-in 2949  df-ss 2956
This theorem is referenced by:  sscond  3105
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