Show me how!

So...
How do you prove:

(A \ B) \ C ⊆ A \ (B \ C) ?

It's pretty fucking obvious

what does \ mean?

\ is the exclusion.
In layman's terms it is to prove that:

(A without B) without C is a subset of A without (B without C)

Venn diagram lol

>>5

While Venn diagrams may be quite nice to look at, they do not fulfill the criteria for a formal proof.

Jeepers, don't you have something constructive to do, just proving shit all day

Take an element a of (A \ B) \ C and show it's also a member of A \ (B \ C)

a E A\B
=> a E A
and a E B'
a E C'

Now for A\(B\C) we have
a E A
and a E (B\C)'
ie a E(BnC')'

so a E An(B'uC) from which we can see  that a E A, a E B' and
a E C' fulfill.

>>6
A proper Venn Diagram does, in fact, satisfy the criteria for a formal proof. It suffices, when proving this, to look at what happens at eight (2^3) possible elements and treat them individually, which is equivalent to what a Venn Diagram would do.

The eight elements are representative of (1) elements not in A, B, or C, (2) elements in A but not B or C, (3) elements in B but not A or C, ... and so on, much as the regions of a Venn Diagram are.