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Theorem ssdif0im 3366
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 826 . . . 4 ((𝑥𝐴𝑥𝐵) → ¬ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
2 eldif 3022 . . . 4 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
31, 2sylnibr 640 . . 3 ((𝑥𝐴𝑥𝐵) → ¬ 𝑥 ∈ (𝐴𝐵))
43alimi 1396 . 2 (∀𝑥(𝑥𝐴𝑥𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴𝐵))
5 dfss2 3028 . 2 (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
6 eq0 3320 . 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 1294   = wceq 1296  wcel 1445  cdif 3010  wss 3013  c0 3302
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 582  ax-in2 583  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077
This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-v 2635  df-dif 3015  df-in 3019  df-ss 3026  df-nul 3303
This theorem is referenced by:  vdif0im  3367  difrab0eqim  3368  difid  3370  difin0  3375
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