Doing a proof for sets by element method and chain of equalities

[lookingforhelp]

I've gotten started on doing this proof, but I'm stuck both ways. I would really appreciate some help with these. Thank you!

For all sets A, B, and C, (A - B) ∪ (B - C) = (A ∪ B) - (B ∩ C)

Proof: by element method
Let x∈(A - B) ∪ (B - C)
By definition of union x∈ A - B or x∈ B - C
x∈A and x∉B or x∈B and x∉C
...
x∈(A ∪ B) - (B ∩ C)
(A - B) ∪ (B - C)⊆(A ∪ B) - (B ∩ C)

Let x∈(A ∪ B) - (B ∩ C)
....
x∈(A - B) ∪ (B - C)
(A - B) ∪ (B - C)⊇(A ∪ B) - (B ∩ C)

Proof: by chain of equalities
(A - B) ∪ (B - C) = (A ∩ B') ∪ (B ∩ C') [set difference] =
....
= (A ∪ B) - (B ∩ C)

 

[daon2]

I've gotten started on doing this proof, but I'm stuck both ways. I would really appreciate some help with these. Thank you!

For all sets A, B, and C, (A - B) ∪ (B - C) = (A ∪ B) - (B ∩ C)

Proof: by element method
Let x∈(A - B) ∪ (B - C)
By definition of union x∈ A - B or x∈ B - C
x∈A and x∉B or x∈B and x∉C

So, case 1 is x belongs to A but not B. Then x belongs to A u B (since its in A) and x does not belong to BnC (since it does not belong to B), so x belongs to (AuB)-(BnC) Try case 2 now.

∈(A ∪ B) - (B ∩ C)
(A - B) ∪ (B - C)⊆(A ∪ B) - (B ∩ C)

Let x∈(A ∪ B) - (B ∩ C)

So x belongs to AuB but x does not belong to BnC, i.e. x belongs to A or x belongs to B, but x does not belong to both B and C

case 1: x belongs to only A and not B.. then x belongs to (A-B)
case 2: x belongs to B (and possibly A).. then x belongs to (B-C)

x∈(A - B) ∪ (B - C)
(A - B) ∪ (B - C)⊇(A ∪ B) - (B ∩ C)

Proof: by chain of equalities
(A - B) ∪ (B - C) = (A ∩ B') ∪ (B ∩ C') [set difference] =
....
= (A ∪ B) - (B ∩ C)