Theorem ddifss 3365

Description: Double complement under universal class. In classical logic (or given an additional hypothesis, as in ddifnel 3258), 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 3169	. 2 ⊢ 𝐴 ⊆ V
2		ssddif 3361	. 2 ⊢ (𝐴 ⊆ V ↔ 𝐴 ⊆ (V ∖ (V ∖ 𝐴)))
3	1, 2	mpbi 144	1 ⊢ 𝐴 ⊆ (V ∖ (V ∖ 𝐴))

Syntax hints:  Vcvv 2730   ∖ cdif 3118   ⊆ wss 3121

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

This theorem is referenced by:  ssindif0im  3474  difdifdirss  3499