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Theorem ssdif0im 3502
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 689 . . . 4 ((𝑥𝐴𝑥𝐵) → ¬ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
2 eldif 3153 . . . 4 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
31, 2sylnibr 678 . . 3 ((𝑥𝐴𝑥𝐵) → ¬ 𝑥 ∈ (𝐴𝐵))
43alimi 1466 . 2 (∀𝑥(𝑥𝐴𝑥𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴𝐵))
5 dfss2 3159 . 2 (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
6 eq0 3456 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴𝐵))
74, 5, 63imtr4i 201 1 (𝐴𝐵 → (𝐴𝐵) = ∅)
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 104  wal 1362   = wceq 1364  wcel 2160  cdif 3141  wss 3144  c0 3437
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2171
This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-nfc 2321  df-v 2754  df-dif 3146  df-in 3150  df-ss 3157  df-nul 3438
This theorem is referenced by:  vdif0im  3503  difrab0eqim  3504  difid  3506  difin0  3511
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