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Theorem ssdif0im 3427
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 677 . . . 4 ((𝑥𝐴𝑥𝐵) → ¬ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
2 eldif 3080 . . . 4 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
31, 2sylnibr 666 . . 3 ((𝑥𝐴𝑥𝐵) → ¬ 𝑥 ∈ (𝐴𝐵))
43alimi 1431 . 2 (∀𝑥(𝑥𝐴𝑥𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴𝐵))
5 dfss2 3086 . 2 (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
6 eq0 3381 . 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 1329   = wceq 1331  wcel 1480  cdif 3068  wss 3071  c0 3363
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121
This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-dif 3073  df-in 3077  df-ss 3084  df-nul 3364
This theorem is referenced by:  vdif0im  3428  difrab0eqim  3429  difid  3431  difin0  3436
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