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
Thanks for your help.
To prove the given statements, we will use basic set operations and properties of sets. Let's take each statement one by one.
Statement 1: For all sets A, B, and C, if A - (B ∩ C) = ∅, then A - C ⊆ B.
Proof:
1. Assume A - (B ∩ C) = ∅.
2. Suppose x is an arbitrary element of A - C.
3. By the definition of set difference, x ∈ A and x ∉ C.
4. Since x ∉ C, x ∉ (B ∩ C).
5. From the assumption in step 1, A - (B ∩ C) = ∅, which implies that A ⊆ (B ∩ C).
6. Combining step 3 and step 5, we have x ∈ (B ∩ C).
7. From step 4 and step 6, we have a contradiction, as x cannot simultaneously be in (B ∩ C) and not in (B ∩ C).
8. Therefore, the assumption in step 2 is false, and A - C ⊆ B.
Statement 2: For all sets A, B, and C, if A ⊆ B ∪ C and B ⊆ C ∪ A, then A ∆ B ⊆ C.
Proof:
1. Assume A ⊆ B ∪ C and B ⊆ C ∪ A.
2. Let x be an arbitrary element in A ∆ B.
3. By the definition of symmetric difference, x ∈ (A - B) ∪ (B - A).
4. If x ∈ (A - B), then x ∈ A and x ∉ B.
5. By the assumption in step 1, A ⊆ B ∪ C, we have x ∈ B ∪ C.
6. If x ∈ B, then x ∈ (B - A), which contradicts the assumption in step 3.
7. Therefore, by step 4, we can conclude that x ∈ C.
8. If x ∈ (B - A), then x ∈ B and x ∉ A.
9. By the assumption in step 1, B ⊆ C ∪ A, we have x ∈ C ∪ A.
10. If x ∈ C, then x ∈ (A - B), which contradicts the assumption in step 3.
11. Therefore, by step 8, we can conclude that x ∈ C.
12. Combining steps 7 and 11, we have x ∈ C.
13. Since x was an arbitrary element in A ∆ B, we can conclude that A ∆ B ⊆ C.
Statement 3: For all sets A, B, and C, if A ∆ B = A ∆ C, then B ⊆ C.
Proof:
1. Assume A ∆ B = A ∆ C.
2. Take an arbitrary element x in B.
3. We aim to show that x ∈ C.
4. By the definition of symmetric difference, x ∈ (A - B) ∪ (B - A).
5. Since x ∈ B, x ∈ (B - A).
6. By assumption in step 1, A ∆ B = A ∆ C, we have (B - A) = (C - A).
7. Therefore, by step 5 and step 6, x ∈ C - A.
8. By the definition of set difference, x ∉ A.
9. Since x ∈ C - A and x ∉ A, we have x ∈ C.
10. Since x was an arbitrary element in B, we can conclude that B ⊆ C.
By using these proofs, we have demonstrated the validity of the given statements.