Let R be the statement : for all sets A, B and C, if A⊆B∪C and B ⊆C∪A then AΔB = C.
Q: Is R true for all sets A, B and C? Prove answer
Let S be the statement: for all sets A, B, and C, if AΔB = AΔC then B⊆C.
Q: Is S true? prove answer
Thanks for your help.
To prove whether the statements R and S are true or false, we need to carefully analyze the definitions and properties involved. Let's begin with statement R:
R: For all sets A, B, and C, if A ⊆ B∪C and B ⊆ C∪A, then AΔB = C.
To prove that R is true, we need to show that if the given conditions hold, then AΔB is indeed equal to C.
First, let's break down the definition of AΔB:
AΔB = (A - B) ∪ (B - A)
Now, consider the given conditions:
1. A ⊆ B∪C: This means that every element in A is either in B or in C (or both).
2. B ⊆ C∪A: This means that every element in B is either in C or in A (or both).
Now, let's reason step by step:
Step 1: Let x be any element in AΔB.
By the definition of symmetric difference, x must either be in A but not in B, or in B but not in A.
Case 1: x ∈ A and x ∉ B.
Since A ⊆ B∪C, x must either be in B or in C. However, x cannot be in B since x ∉ B. Therefore, x ∈ C.
Case 2: x ∈ B and x ∉ A.
Since B ⊆ C∪A, x must either be in C or in A. However, x cannot be in A since x ∉ A. Therefore, x ∈ C.
In both cases, we have shown that any element x in AΔB is also in C.
Step 2: Let x be any element in C.
By the definition of set union, x must either be in B or in A (or both). In either case, x would not be in A - B or B - A since it is already in both A and B. Thus, x ∉ AΔB.
Step 1 shows that any element in AΔB is also in C, and step 2 shows that any element in C is not in AΔB. Therefore, we can conclude that AΔB = C.
Hence, we have proved that statement R is true for all sets A, B, and C.
Now, let's move on to statement S:
S: For all sets A, B, and C, if AΔB = AΔC, then B ⊆ C.
To prove that S is true, we need to show that if the given condition holds, then B is indeed a subset of C.
Let's reason step by step:
Step 1: Assume that AΔB = AΔC.
Step 2: Let x be any element in B.
By the definition of symmetric difference, x must either be in B but not in A, or in A but not in B.
Case 1: x ∈ B and x ∉ A.
Since x ∈ B and AΔB = AΔC, we can conclude that x ∈ AΔC. By the definition of symmetric difference and set union, x must either be in C but not in A, or in A but not in C.
Case 1.1: x ∈ C but x ∉ A.
In this case, x is in both B and C, and we can conclude that x ∈ C.
Case 1.2: x ∈ A but x ∉ C.
Since x ∈ A and AΔB = AΔC, we can conclude that x ∈ AΔB. By the definition of symmetric difference and set union, x must either be in B but not in A, or in A but not in B. However, we already assumed that x ∉ A, so x must be in B but not in A. Therefore, x ∈ B - A, which contradicts our assumption in case 1.
From case 1.1, we have shown that any element x in B is also in C.
Therefore, we can conclude that B ⊆ C.
Hence, we have proved that statement S is true for all sets A, B, and C.
In summary, statement R is true for all sets A, B, and C, and statement S is also true for all sets A, B, and C.