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Theorem ssdif0im 3478
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 683 . . . 4 ((𝑥𝐴𝑥𝐵) → ¬ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
2 eldif 3130 . . . 4 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
31, 2sylnibr 672 . . 3 ((𝑥𝐴𝑥𝐵) → ¬ 𝑥 ∈ (𝐴𝐵))
43alimi 1448 . 2 (∀𝑥(𝑥𝐴𝑥𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴𝐵))
5 dfss2 3136 . 2 (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
6 eq0 3432 . 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 1346   = wceq 1348  wcel 2141  cdif 3118  wss 3121  c0 3414
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 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-dif 3123  df-in 3127  df-ss 3134  df-nul 3415
This theorem is referenced by:  vdif0im  3479  difrab0eqim  3480  difid  3482  difin0  3487
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