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Theorem disj4im 3253
 Description: A consequence of two classes being disjoint. In classical logic this would be a biconditional. (Contributed by Jim Kingdon, 2-Aug-2018.)
Assertion
Ref Expression
disj4im ((AB) = ∅ → ¬ (AB) ⊊ A)

Proof of Theorem disj4im
StepHypRef Expression
1 disj3 3249 . . 3 ((AB) = ∅ ↔ A = (AB))
2 eqcom 2024 . . 3 (A = (AB) ↔ (AB) = A)
31, 2bitri 173 . 2 ((AB) = ∅ ↔ (AB) = A)
4 dfpss2 3006 . . . 4 ((AB) ⊊ A ↔ ((AB) ⊆ A ¬ (AB) = A))
54simprbi 260 . . 3 ((AB) ⊊ A → ¬ (AB) = A)
65con2i 545 . 2 ((AB) = A → ¬ (AB) ⊊ A)
73, 6sylbi 114 1 ((AB) = ∅ → ¬ (AB) ⊊ A)
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   = wceq 1228   ∖ cdif 2891   ∩ cin 2893   ⊆ wss 2894   ⊊ wpss 2895  ∅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-ne 2188  df-ral 2289  df-v 2537  df-dif 2897  df-in 2901  df-pss 2910  df-nul 3202 This theorem is referenced by: (None)
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