Theorem inssddif 3450

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 3429	. . 3 ⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴
2		ssddif 3443	. . 3 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐴 ↔ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ (𝐴 ∩ 𝐵))))
3	1, 2	mpbi 145	. 2 ⊢ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ (𝐴 ∩ 𝐵)))
4		difin 3446	. . 3 ⊢ (𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ 𝐵)
5	4	difeq2i 3324	. 2 ⊢ (𝐴 ∖ (𝐴 ∖ (𝐴 ∩ 𝐵))) = (𝐴 ∖ (𝐴 ∖ 𝐵))
6	3, 5	sseqtri 3262	1 ⊢ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ 𝐵))

Syntax hints:   ∖ cdif 3198   ∩ cin 3200   ⊆ wss 3201

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 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-fal 1404  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-ral 2516  df-rab 2520  df-v 2805  df-dif 3203  df-in 3207  df-ss 3214

This theorem is referenced by: (None)