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 notation and basic principles of set theory. Let's start with the first statement:
1. For all sets A, B, and C, if A - (B ∩ C) = ∅, then A - C ⊆ B.
Proof:
To begin, let's assume that A - (B ∩ C) = ∅.
The difference A - (B ∩ C) represents the elements that are in set A but not in the intersection of sets B and C. Since this difference is empty (represented by ∅), it means that every element in set A is also present in the intersection of sets B and C.
Now, let's consider the set A - C. This set contains elements that are in set A but not in set C. We want to prove that A - C is a subset of set B.
Suppose there exists an element x ∈ A - C. This implies that x is an element of set A but not an element of set C.
Since every element in set A is also in the intersection of sets B and C (as stated above), it follows that x ∈ (B ∩ C). But since x ∉ C, it implies that x ∈ B.
Therefore, for any element x ∈ A - C, it is also an element of B. Hence, A - C ⊆ B.
Next, let's move to the second statement:
2. For all sets A, B, and C, if A ⊆ B ∪ C and B ⊆ C ∪ A, then A ∆ B ⊆ C.
Proof:
Assume that A ⊆ B ∪ C and B ⊆ C ∪ A.
The symmetric difference A ∆ B can be defined as (A ∪ B) - (A ∩ B). Let's calculate this expression:
(A ∪ B) - (A ∩ B) = (A ∪ B) ∩ (A ∩ B)'
By De Morgan's Laws, we can rewrite this as:
(A ∪ B) - (A ∩ B) = (A ∩ B') ∪ (B ∩ A')
Now, we need to show that (A ∩ B') ∪ (B ∩ A') ⊆ C.
Considering (A ∩ B') ∪ (B ∩ A'), we know that for an element x to belong to this set, it must satisfy one of the following conditions:
1. x ∈ A ∩ B'
2. x ∈ B ∩ A'
Let's examine these conditions separately:
1. If x ∈ A ∩ B', it means that x is in both sets A and B' (the complement of B). Since A ⊆ B ∪ C, it follows that every element in A is also in either B or C. However, since x ∉ B, x must be an element of C. Thus, x ∈ C.
2. If x ∈ B ∩ A', it means that x is in both sets B and A' (the complement of A). Similarly, since B ⊆ C ∪ A, every element in B is in either C or A. But since x ∉ A, x must be an element of C. Hence, x ∈ C.
Based on these two conditions, we can conclude that for any element x in (A ∩ B') ∪ (B ∩ A'), x ∈ C.
Therefore, A ∆ B ⊆ C.
Finally, let's move to the third statement:
3. For all sets A, B, and C, if A ∆ B = A ∆ C, then B ⊆ C.
Proof:
Assume A ∆ B = A ∆ C.
By the definition of symmetric difference, we can expand this equation as:
(A ∪ B) - (A ∩ B) = (A ∪ C) - (A ∩ C)
Using the properties of set operations, we can simplify this expression:
(A ∪ B) ∩ (A ∩ C)' = (A ∪ C) ∩ (A ∩ B)'
To prove the desired result, we will show that B ⊆ C by considering an arbitrary element x ∈ B.
Since x ∈ B, it implies that x ∈ (A ∪ B). According to the equation above, this means that x must also belong to (A ∪ C). Therefore, x ∈ (A ∪ C) ∩ (A ∩ B)'.
By De Morgan's Laws, we can rewrite this as:
x ∈ (A ∩ C') ∪ (A' ∩ B')
Since we assume A ∆ B = A ∆ C, it means that this expression simplifies to:
x ∈ (A ∩ C') ∪ (A' ∩ C)
From this, we can conclude that x ∈ C.
Since this is true for any element x ∈ B, it implies that B ⊆ C.
Therefore, if A ∆ B = A ∆ C, then B ⊆ C.
By completing these proofs, we have established the validity of the given statements.