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
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To prove these statements, we can break down the logic step by step.
1. For all sets A, B, and C, if A - (B ∩ C) = ∅, then A - C ⊆ B.
To prove this, we need to show that every element in A - (B ∩ C) is also in B.
Let's assume x is an arbitrary element in A - (B ∩ C). This means that x is in A, but not in (B ∩ C). Therefore, x is not in B or x is not in C.
If x is not in C, then x is also in A - C. Therefore, x is in A - C ⊆ B.
If x is not in B, then x is in A - (B ∩ C), which contradicts our assumption. Hence, x must be in B.
Since we've shown that every element in A - (B ∩ C) is also in B, we 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.
To prove this, we need to show that every element in A∆B is also in C.
Let's assume x is an arbitrary element in A∆B. This means that x is either in (A - B) or (B - A).
If x is in (A - B), then x is in A but not in B. Since A ⊆ B ∪ C, x must be in (B ∪ C) but not in B. Hence, x must be in C.
If x is in (B - A), then x is in B but not in A. Since B ⊆ C ∪ A, x must be in (C ∪ A) but not in A. Hence, x must be in C.
In both cases, we've shown that every element in A∆B is also in C. Therefore, A∆B ⊆ C.
3. For all sets A, B, and C, if A∆B = A∆C, then B ⊆ C.
To prove this, we need to show that every element in B is also in C.
Let's assume x is an arbitrary element in B. We want to show that x is also in C.
Since A∆B = A∆C, every element in B - A is also in C - A, and every element in A - B is also in A - C.
If x is in B - A, then x is in C - A, which means it is in C but not in A.
If x is in A - B, then x is in A - C, which means it is not in C.
In both cases, x must be in C. Hence, we conclude that B ⊆ C.
By proving these statements step by step, we have demonstrated their validity for any sets A, B, and C.