(I)

(A-B)∪(A∩B)

[x ∈ A or x ∈ A]

x ∈ A --> 1° Case (A-B) (in A)

x ∈ A --> 2° Case (A∩B) (in A)

you determined that x belongs to A, so x is in A (A-B) and A (A∩B) so it shows that (A-B)∪(A∩B) ⊂ A ?

No, I didn't say "x belongs to A, so x is in A (A-B)

**and** A (A∩B)"; I said that if x is in either (A-B)

**or** (A∩B), then x is in A.

But, yes, my conclusion is that (A-B)∪(A∩B) ⊂ A.

(II)

[[x ∈ A]]

(A-B)∪(A∩B)

[x ∈ B or x ∉ B]

x ∈ B --> 1° case (A∩B) (in B)

x ∉ B --> 2° case (A-B) (in -B)

you determined that x belongs to B or x does not belong to B (-B) in (A-B) in(-B) and in (A∩B) in(B) it shows that A ⊂ (A-B)∪(A∩B)? wht, i don't understand ;;(((

Maybe I'm just not understanding your notation here (or above). What do you mean when you write "x is in A (A-B)" and, here, "x does not belong to B (-B) in (A-B) in(-B)"?

This is the case I

*didn't* do for you, so I'm not sure what "you determined" refers to.

If x is in A, then either it is

**not **in B (so it is in A-B), or it

**is **in B (so it is in A∩B). Therefore it is in either A-B or A∩B. Therefore A ⊂ (A-B)∪(A∩B).