Elementary-set theory question, I really need a help with this one please!?

(A\B)\C ⊆ A∖(B∖C)

(A\B)\C = (A\C)\B = A\(B U C)

(A\B)\C = (A∩B’)∩ C’

=(A∩C’) ∩ (B’∩C’) … distribution

=A∩ C’ ∩ (B U C)’ … de Morgan’s law

=A∩(B UC)’ … absorption

=A\(B U C)

Because (B\C) is always smaller than or equal to (B U C) … or (B\C) ⊆ (B U C)

So A\(B U C) is always smaller than or equal to A\(B\C) … or A\(BUC) ⊆ A\(B\C)’

For examples

Ex1

A = {1, 2 ,3, 4,5} B = {2,3} C={3,5}

(A\B)\C = A\(B U C) = {1,4}

A\(B\C) = {1,4,5}

(A\B)\C ⊆ A∖(B∖C)

Ex2

A = {1, 2 ,3, 4,5} B = {2,3} C={4,5}

(A\B)\C = A\(B U C) = {1}

A\(B\C) = {1,4,5}

(A\B)\C ⊆ A∖(B∖C)

Ex3

A = {1, 2 ,3, 4,5} B = {2,3} C={}

(A\B)\C = A\(B U C) = {1,4,5}

A\(B\C) = {1,4,5}

(A\B)\C ⊆ A∖(B∖C)

If you can assume a universe set U (the union of all sets under consideration, or any superset of that) then set algebra makes this easy:

(A\B)\C = (A\B)∩C' = (A ∩ B') ∩ C'

Since both union and intersection are associative, I'll write multiple unions or intersections without parentheses, so (A\B)\C = A ∩ B' ∩ C'

A\(B\C) = A \ (B∩C') = A ∩ (B ∩ C')' = A ∩ (B' ∪ C) .... de Morgan's Law

= (A ∩ B') ∪ (A∩C) .... distributive property

= [(A ∩ B') ∩ (C' ∪ C)] ∪ (A ∩ C) .... C∪C' = U is an identity element for intersection

= (A ∩ B' ∩ C') ∪ (A ∩ B ∩ C) ∪ (A ∩ C) ... distributive again

= (A ∩ B' ∩ C') ∪ (A ∩ C) ... absorption

= [ (A\B)\C ] ∪ (A ∩ C)

This makes (A\B)\C ⊆ A\(B\C), with equality if and only if A∩C is empty. (i.e. when A and C are disjoint)

Without the universe set, I don't see an easier proof than considering which of the 8 cases where an arbitrary element x taken from (A\B)\C is (in/not in) A, B and/or C...and then showing that the possible cases all show that x must also be in A\(B\C).