1. for any sets X and Y, we define the symmetric difference X ∆Y by:

X∆Y = (X - Y) ∪ (Y - X) = (X ∪ Y) - (X ∩ Y)
Prove the following:
-for all sets A,B and C, if A - (B ∩ C) = ∅ then A- C ⊆ B
- for all sets A,B and C, if A ⊆ B ∪ C and B ⊆ C ∪ A then A∆B ⊆ C
- for all sets A,B and C, if A∆B = A∆C then B ⊆C

To prove the given statements, we will use set properties and logical reasoning. Let's go through each statement one by one:

1. For all sets A, B, and C, if A - (B ∩ C) = ∅, then A - C ⊆ B.

Proof:
- Assume A - (B ∩ C) = ∅, which means that A and (B ∩ C) have no elements in common.
- Now, let's consider an element x in A - C (i.e., an element in A but not in C).
- Since A and (B ∩ C) have no elements in common, we can conclude that x is not in B ∩ C.
- Since x is not in B ∩ C, it implies that x is neither in B nor in C.
- Therefore, x must be in A but not in C, which means x is in A - (B ∩ C).
- But we assumed that A - (B ∩ C) = ∅, which leads to a contradiction.
- This contradiction arises because we assumed that an element x in A - C exists, but we found that it is also in A - (B ∩ C), which was assumed to be empty.
- Hence, our assumption must be incorrect, and we can conclude that A - C ⊆ B.

2. For all sets A, B, and C, if A ⊆ B ∪ C and B ⊆ C ∪ A, then A∆B ⊆ C.

Proof:
- Assume A ⊆ B ∪ C and B ⊆ C ∪ A.
- We need to prove that A∆B ⊆ C, which means any element in A∆B is also in C or not in A ∩ B.
- Let's consider an element x in A∆B (i.e., x is in either A or B but not in both).
- Case 1: If x is in A, then it is not in B because A and B are disjoint in A∆B.
- Since A ⊆ B ∪ C, and x is in A but not in B, x must be in C.
- Therefore, x is in C, and A∆B ⊆ C in this case.
- Case 2: If x is in B, then it is not in A because A and B are disjoint in A∆B.
- Since B ⊆ C ∪ A, and x is in B but not in A, x must be in C.
- Therefore, x is in C, and A∆B ⊆ C in this case.
- Thus, in both cases, we have shown that any element x in A∆B is also in C or not in A ∩ B.
- Hence, A∆B ⊆ C.

3. For all sets A, B, and C, if A∆B = A∆C, then B ⊆ C.

Proof:
- Assume A∆B = A∆C.
- We need to prove that B ⊆ C, which means any element in B is also in C.
- Let's consider an element x in B.
- Since x is in B, it implies that either x is in A or x is in B but not in A (B - A).
- If x is in A, then it cannot be in A∆B because A∆B only contains elements that are in A but not in B.
- As A∆B = A∆C, x cannot be in A∆C either.
- Hence, x must be in C.
- If x is in B but not in A (B - A), then it is also in A∆B.
- But A∆B = A∆C, so x must be in A∆C as well.
- Since x is in A∆C, it means x is in A but not in C.
- This contradicts our assumption that x is in B and not in A∆C (which implies x is not in A but in C).
- Therefore, our assumption that x is in B leads to a contradiction.
- Hence, x cannot be in B, and we can conclude that B ⊆ C.

By proving all three statements, we have verified the given statements for any sets A, B, and C.