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Theorem disj4im 3270
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 3266 . . 3 ((AB) = ∅ ↔ A = (AB))
2 eqcom 2039 . . 3 (A = (AB) ↔ (AB) = A)
31, 2bitri 173 . 2 ((AB) = ∅ ↔ (AB) = A)
4 dfpss2 3023 . . . 4 ((AB) ⊊ A ↔ ((AB) ⊆ A ¬ (AB) = A))
54simprbi 260 . . 3 ((AB) ⊊ A → ¬ (AB) = A)
65con2i 557 . 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 1242  cdif 2908  cin 2910  wss 2911  wpss 2912  c0 3218
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 544  ax-in2 545  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ne 2203  df-ral 2305  df-v 2553  df-dif 2914  df-in 2918  df-pss 2927  df-nul 3219
This theorem is referenced by: (None)
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