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Theorem sscon 3400
Description: Contraposition law for subsets. Exercise 15 of [TakeutiZaring] p. 22. (Contributed by NM, 22-Mar-1998.)
Assertion
Ref Expression
sscon (A B → (C B) (C A))

Proof of Theorem sscon
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 ssel 3267 . . . . 5 (A B → (x Ax B))
21con3d 125 . . . 4 (A B → (¬ x B → ¬ x A))
32anim2d 548 . . 3 (A B → ((x C ¬ x B) → (x C ¬ x A)))
4 eldif 3221 . . 3 (x (C B) ↔ (x C ¬ x B))
5 eldif 3221 . . 3 (x (C A) ↔ (x C ¬ x A))
63, 4, 53imtr4g 261 . 2 (A B → (x (C B) → x (C A)))
76ssrdv 3278 1 (A B → (C B) (C A))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   wa 358   wcel 1710   cdif 3206   wss 3257
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215  df-ss 3259
This theorem is referenced by:  sscond  3403
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