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Theorem ssdif0im 3397
Description: Subclass implies empty difference. One direction of Exercise 7 of [TakeutiZaring] p. 22. In classical logic this would be an equivalence. (Contributed by Jim Kingdon, 2-Aug-2018.)
Assertion
Ref Expression
ssdif0im (𝐴𝐵 → (𝐴𝐵) = ∅)

Proof of Theorem ssdif0im
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 imanim 662 . . . 4 ((𝑥𝐴𝑥𝐵) → ¬ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
2 eldif 3050 . . . 4 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
31, 2sylnibr 651 . . 3 ((𝑥𝐴𝑥𝐵) → ¬ 𝑥 ∈ (𝐴𝐵))
43alimi 1416 . 2 (∀𝑥(𝑥𝐴𝑥𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴𝐵))
5 dfss2 3056 . 2 (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
6 eq0 3351 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴𝐵))
74, 5, 63imtr4i 200 1 (𝐴𝐵 → (𝐴𝐵) = ∅)
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 103  wal 1314   = wceq 1316  wcel 1465  cdif 3038  wss 3041  c0 3333
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 588  ax-in2 589  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-v 2662  df-dif 3043  df-in 3047  df-ss 3054  df-nul 3334
This theorem is referenced by:  vdif0im  3398  difrab0eqim  3399  difid  3401  difin0  3406
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