Theorem inssddif 3205

Description: Intersection of two classes and class difference. In classical logic, such as Exercise 4.10(q) of [Mendelson] p. 231, this is an equality rather than subset. (Contributed by Jim Kingdon, 26-Jul-2018.)

Assertion

Ref	Expression
inssddif	⊢ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ 𝐵))

Proof of Theorem inssddif

Step	Hyp	Ref	Expression
1		inss1 3186	. . 3 ⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴
2		ssddif 3198	. . 3 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐴 ↔ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ (𝐴 ∩ 𝐵))))
3	1, 2	mpbi 143	. 2 ⊢ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ (𝐴 ∩ 𝐵)))
4		difin 3201	. . 3 ⊢ (𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ 𝐵)
5	4	difeq2i 3087	. 2 ⊢ (𝐴 ∖ (𝐴 ∖ (𝐴 ∩ 𝐵))) = (𝐴 ∖ (𝐴 ∖ 𝐵))
6	3, 5	sseqtri 3031	1 ⊢ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ 𝐵))

Syntax hints:   ∖ cdif 2970   ∩ cin 2972   ⊆ wss 2973

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 576  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-fal 1290  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-rab 2357  df-v 2603  df-dif 2975  df-in 2979  df-ss 2986

This theorem is referenced by: (None)