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Theorem ssdif0im 3263
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 778 . . . 4 ((x Ax B) → ¬ (x A ¬ x B))
2 eldif 2904 . . . 4 (x (AB) ↔ (x A ¬ x B))
31, 2sylnibr 589 . . 3 ((x Ax B) → ¬ x (AB))
43alimi 1324 . 2 (x(x Ax B) → x ¬ x (AB))
5 dfss2 2911 . 2 (ABx(x Ax B))
6 eq0 3216 . 2 ((AB) = ∅ ↔ x ¬ x (AB))
74, 5, 63imtr4i 190 1 (AB → (AB) = ∅)
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4   wa 97  wal 1226   = wceq 1228   wcel 1374  cdif 2891  wss 2894  c0 3201
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 532  ax-in2 533  ax-io 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-10 1377  ax-11 1378  ax-i12 1379  ax-bnd 1380  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004
This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-v 2537  df-dif 2897  df-in 2901  df-ss 2908  df-nul 3202
This theorem is referenced by:  vdif0im  3264  difrab0eqim  3265  difid  3269  difin0  3274
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