Theorem inssddif 3400

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 3379	. . 3 ⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴
2		ssddif 3393	. . 3 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐴 ↔ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ (𝐴 ∩ 𝐵))))
3	1, 2	mpbi 145	. 2 ⊢ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ (𝐴 ∩ 𝐵)))
4		difin 3396	. . 3 ⊢ (𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ 𝐵)
5	4	difeq2i 3274	. 2 ⊢ (𝐴 ∖ (𝐴 ∖ (𝐴 ∩ 𝐵))) = (𝐴 ∖ (𝐴 ∖ 𝐵))
6	3, 5	sseqtri 3213	1 ⊢ (𝐴 ∩ 𝐵) ⊆ (𝐴 ∖ (𝐴 ∖ 𝐵))

Syntax hints:   ∖ cdif 3150   ∩ cin 3152   ⊆ wss 3153

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-fal 1370  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ral 2477  df-rab 2481  df-v 2762  df-dif 3155  df-in 3159  df-ss 3166

This theorem is referenced by: (None)