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Theorem sscon 3210
 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 3091 . . . . 5 (𝐴𝐵 → (𝑥𝐴𝑥𝐵))
21con3d 620 . . . 4 (𝐴𝐵 → (¬ 𝑥𝐵 → ¬ 𝑥𝐴))
32anim2d 335 . . 3 (𝐴𝐵 → ((𝑥𝐶 ∧ ¬ 𝑥𝐵) → (𝑥𝐶 ∧ ¬ 𝑥𝐴)))
4 eldif 3080 . . 3 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
5 eldif 3080 . . 3 (𝑥 ∈ (𝐶𝐴) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐴))
63, 4, 53imtr4g 204 . 2 (𝐴𝐵 → (𝑥 ∈ (𝐶𝐵) → 𝑥 ∈ (𝐶𝐴)))
76ssrdv 3103 1 (𝐴𝐵 → (𝐶𝐵) ⊆ (𝐶𝐴))
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ∈ wcel 1480   ∖ cdif 3068   ⊆ wss 3071 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121 This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-dif 3073  df-in 3077  df-ss 3084 This theorem is referenced by:  sscond  3213  sbthlem1  6845  sbthlem2  6846
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