Problem 2 (15 pts.) Empecied Time: 1-2 hours. Similar to: Tutorial '7 Problems I 3 2, Claim in Mathematical Induction lecture. Type of Practice: Read, Parse and Deconstruct Given Proofs. This problem asks you to read a chatGPT proof and improve upon it. It also asks you to construct a proof by mathematical induc- tion, in the format used in this course (eg. Tstep process from lecture, using the \"column format\" in the inductive step, etc.) Claim. Let n 2 2, and let A1, A2, . . . ,An be sets from some universal set U. For all n 2 2, 1!. _ Um=m. j=1 i=1 Recall from Set Operations, Pt. 2: n n UAg=A1UA2U'-'UAH and Ag=A1nA2n-HnAn 1 Consider the following chatGPT proof of the result (where ~A = Z.) To prove this statement by mathematical induction. you need to show that it holds true tor a base case and then prove that if it holds true for some arbitrary positive integer n. it also holds to: n+1. Base Case (n = 2): For the base case. let's consider n = 2: We want to prove that: ~(A1I_IA2)= ~A1 n -A2t We'll start by using De Morgan's Laws to expand both sides at the equation: -(A1I_IA2)= -A1 n -A2. So. the base case is established. Inductive Step: Now, assume that the statement holds for some arbitrary positive integer k > 2. That is, we assume: -(A1 u A2 U .. U Ak) = -Ain -A20...n-Ak. We want to prove that it also holds for k+1: (A1 u A2 u ... U Ak u Ak+1) = -A1n-A20..n-Akn -Ak+1. To do this, we'll use the assumption and De Morgan's Law for the intersection of two sets: -(A U B) = -An-B. First, we'll apply De Morgan's Law for the intersection of two sets to the right-hand side of our assumption: -(A1 u A2 u .. U Ak) = -A1n-A20... n-Ak. Now, we'll rewrite the expression on the left-hand side of our inductive step using the right- hand side of the assumption and the De Morgan's Law for the intersection of two sets: -(A1 u A2 u ... U Ak u Ak+1) = =(A1 u A2 U... U AK) n -Ak+1. Using the inductive assumption, we can substitute -(A1 u A2 u ... u Ak) with -A10 -A20 ._n -Ak: -(A1 u A2 U .. U Ak u Ak+1) = (-A1 n -A2 0... n-Ak) n-Ak+1. Now, we can use the associative property of set intersection to group the terms on the right- hand side: -(A1 u A2 U .. U Aku Ak+1) = -A1n-A20..n-Akn -Ak+1. This completes the inductive step. By induction, we've shown that if the statement holds for n = 2 (the base case) and if it holds for some arbitrary positive integer k, then it also holds for k+1. Therefore, the statement is proved for all n > 2 by mathematical induction. 6Fill in the blanks below with a proof by mathematical induction of the claim, using this chatGPT proof as a guide. As always, use reasons from the approved list. 1. chatGPT's proof is mostly correct, but is not as structured as the proof guide below. This makes the chatGPT proof harder to follow. Your task is to sort it out and then rewrite it with our structure so that it is easy to follow. Do not include incorrect parts. 2. chatGPT does not dene a predicate; you must do so and then use it correctly in your proof. 3. chatGPT uses ellipses (e.g. A1 UAg - - - UAn) but you must not have any ellipses. You should use U32? and 03:? notation wherever possible. (The \"i = \" part gets printed slightly to the side when the big union / big intersection is not on its own line.) 4. chatGPT uses ~A to represent the complement of set A. Please use the standard 2 instead. Step 0. For all n, 2 , we want to show that Step 1. For any n 2 , let P(n) be the property that We want to show that P(n) is true for all n 2 . Step 2. As a base case, consider when n = . We will show that P( ) is true: that is, that . Fortunately, Step 3. For the induction hypothesis, suppose (hypothetically) that P(lc) were true for some fixed In: 2 . That is, suppose that Step 4. Now we prove that PU: + l) is true, using the (hypothetical) induction assumption that P(k) is true. That is, we prove that Step 5. The proof that PU: + 1) is true (given that PUC) is true) is as follows: left-hand side of PU: + 1) = column format, use IH somewhere = right-hand side of PU: + 1). Therefore we have shown that if P(k) is true, then PUC + 1) is also true, for any It 2 7 Step 6. The steps above have shown that for any I: 2 , if P{k) is true, then PU: + 1) is also true. Combined with the base case, which shows that P( ) is true1 we have shown that for all n 2 , P(n) is true, as desired