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Theorem neldifsnd 3624
Description: 𝐴 is not in (𝐵 ∖ {𝐴}). Deduction form. (Contributed by David Moews, 1-May-2017.)
Assertion
Ref Expression
neldifsnd (𝜑 → ¬ 𝐴 ∈ (𝐵 ∖ {𝐴}))

Proof of Theorem neldifsnd
StepHypRef Expression
1 neldifsn 3623 . 2 ¬ 𝐴 ∈ (𝐵 ∖ {𝐴})
21a1i 9 1 (𝜑 → ¬ 𝐴 ∈ (𝐵 ∖ {𝐴}))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wcel 1465  cdif 3038  {csn 3497
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 588  ax-in2 589  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-ne 2286  df-v 2662  df-dif 3043  df-sn 3503
This theorem is referenced by:  difsnb  3633  frirrg  4242  elirr  4426
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