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Theorem ddifss 3413
Description: Double complement under universal class. In classical logic (or given an additional hypothesis, as in ddifnel 3306), this is equality rather than subset. (Contributed by Jim Kingdon, 24-Jul-2018.)
Assertion
Ref Expression
ddifss 𝐴 ⊆ (V ∖ (V ∖ 𝐴))

Proof of Theorem ddifss
StepHypRef Expression
1 ssv 3217 . 2 𝐴 ⊆ V
2 ssddif 3409 . 2 (𝐴 ⊆ V ↔ 𝐴 ⊆ (V ∖ (V ∖ 𝐴)))
31, 2mpbi 145 1 𝐴 ⊆ (V ∖ (V ∖ 𝐴))
Colors of variables: wff set class
Syntax hints:  Vcvv 2773  cdif 3165  wss 3168
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188
This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-dif 3170  df-in 3174  df-ss 3181
This theorem is referenced by:  ssindif0im  3522  difdifdirss  3547
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