The answer is b because 2S + L is equal to L + 2S.
Answer:
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Answer:
We have to prove,
(A \ B) ∪ ( B \ A ) = (A U B) \ (B ∩ A).
Suppose,
x ∈ (A \ B) ∪ ( B \ A ), where x is an arbitrary,
⇒ x ∈ A \ B or x ∈ B \ A
⇒ x ∈ A and x ∉ B or x ∈ B and x ∉ A
⇒ x ∈ A or x ∈ B and x ∉ B and x ∉ A
⇒ x ∈ A ∪ B and x ∉ B ∩ A
⇒ x ∈ ( A ∪ B ) \ ( B ∩ A )
Conversely,
Suppose,
y ∈ ( A ∪ B ) \ ( B ∩ A ), where, y is an arbitrary.
⇒ y ∈ A ∪ B and x ∉ B ∩ A
⇒ y ∈ A or y ∈ B and y ∉ B or y ∉ A
⇒ y ∈ A and y ∉ B or y ∈ B and y ∉ A
⇒ y ∈ A \ B or y ∈ B \ A
⇒ y ∈ ( A \ B ) ∪ ( B \ A )
Hence, proved......
Answer:
(-1.5,0.5)
Step-by-step explanation:
In the attached file