A subset T of the integers is defined recursively as follows: Base case: 2 ∈ T Recursive rule: if k ∈ T, then k + 5 ∈ T This problem asks you to prove that T is exactly the set of integers that can be expressed as 5m+2, where m is a non-negative integer. In other words, you will prove that x ∈ T if and only if x = 5m+2, for some non-negative integer m. The two directions of the "if and only if" are proven separately. (a) Use structural induction to prove that if k ∈ T, then k = 5m + 2, for some non-negative integer m.

Step-by-step explanation:

We must prove the statement for the base case, and then state our induction hypothesis.

For the base case, note that 2 = 5*0 +2 , so since 0 is a non-negative integer, it is true for 0.

We want to see that if k is in T and has a the property k= 5m+2 for an non negative integer, then the next element of T in the list, i.e k+5 has also this property.

Suppose that for an integer k in T we have an non negative integer m such that k = 5m+2. We have that k+5 = 5m+2 +5 = 5(m+1)+2. So, since m+1 is also a non negative integer, we have proved the claim using structural induction.

21 i watched a 3 hr vid explaining it

step-by-step explanation:

du what else would it be

brainliest

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!