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 these statements, we will use basic set operations and logical reasoning. Let's proceed with each proof step by step:

1. Proving if A - (B ∩ C) = ∅, then A - C ⊆ B:
Assume A - (B ∩ C) = ∅. This means that there are no elements in A that are not in the intersection of B and C. We want to prove that in this case, A - C (the elements in A that are not in C) is a subset of B.

Let x be an arbitrary element in A - C. This means that x is in A but not in C. Since A - (B ∩ C) = ∅, x must also be in B ∩ C since there are no elements in A that are not in B ∩ C.

As x is in both B ∩ C and A, we can conclude that x is in B. Therefore, A - C is a subset of B.

2. Proving if A ⊆ B ∪ C and B ⊆ C ∪ A, then A∆B ⊆ C:
Assume A ⊆ B ∪ C and B ⊆ C ∪ A. We want to prove that in this case, A∆B ⊆ C.

Let x be an arbitrary element in A∆B. This means that x is either in (A - B) or (B - A).

- If x ∈ (A - B): This means that x is in A, but not in B. Since A ⊆ B ∪ C, x must also be in C. Thus, x ∈ (A - B) ⊆ C.

- If x ∈ (B - A): This means that x is in B, but not in A. Since B ⊆ C ∪ A, x must also be in C. Thus, x ∈ (B - A) ⊆ C.

In both cases, we have shown that x is in C. Therefore, A∆B ⊆ C.

3. Proving if A∆B = A∆C, then B ⊆ C:
Assume A∆B = A∆C. We want to prove that in this case, B ⊆ C.

Let x be an arbitrary element in B. We need to show that x is also in C. We will do this by proving that x is in B - A and B - C (from the definition of symmetric difference).

- If x ∈ B - A: This means that x is in B but not in A. Since A∆B = A∆C, x must also be in C but not in A. Therefore, x is in C - A, which implies x is in C.

- Similarly, if x ∈ B - C: This means that x is in B but not in C. Since A∆B = A∆C, x must also be in A but not in C. Therefore, x is in A - C, which implies x is in A, making x also in B.

In both cases, x is in C. Therefore, B ⊆ C.

These proofs demonstrate the validity of the given statements.