Theorem ddifss 3458

Description: Double complement under universal class. In classical logic (or given an additional hypothesis, as in ddifnel 3349), 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

Step	Hyp	Ref	Expression
1		ssv 3259	. 2 ⊢ 𝐴 ⊆ V
2		ssddif 3454	. 2 ⊢ (𝐴 ⊆ V ↔ 𝐴 ⊆ (V ∖ (V ∖ 𝐴)))
3	1, 2	mpbi 145	1 ⊢ 𝐴 ⊆ (V ∖ (V ∖ 𝐴))

Syntax hints:  Vcvv 2812   ∖ cdif 3207   ⊆ wss 3210

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 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-v 2814  df-dif 3212  df-in 3216  df-ss 3223

This theorem is referenced by:  ssindif0im  3567  difdifdirss  3593