If true, prove. If false, provide just a counterexample:

a. For every function f from nonnegative integers into nonnegative reals,
o( f ) = O( f ) −Θ( f ) . (Here, “-“ denotes the set difference: A – B consists of
elements in A that are not in B.).

To prove or disprove the statement, let's break it down.

The statement says that for every function f from nonnegative integers into nonnegative reals, the set of functions in o(f) is equal to the set difference of O(f) and Θ(f).

To prove the statement, we need to show that for every function f, the set of functions in o(f) is equal to O(f) - Θ(f).

To understand what these notations mean, let's define them:

- O(f) represents the set of functions that upper bound f asymptotically. It means that for every function g in O(f), there exist positive constants c and n0 such that for all n ≥ n0, g(n) ≤ c * f(n). In other words, g grows no faster than f for large enough input sizes.

- Θ(f) represents the set of functions that both upper and lower bound f asymptotically. It means that for every function g in Θ(f), there exist positive constants c1, c2, and n0 such that for all n ≥ n0, c1 * f(n) ≤ g(n) ≤ c2 * f(n). In other words, g grows at the same rate as f for large enough input sizes.

- o(f) represents the set of functions that strictly grow slower than f. It means that for every function g in o(f), for all positive constants c, there exists a positive constant n0 such that for all n ≥ n0, g(n) < c * f(n). In other words, g grows slower than f for large enough input sizes.

Now, let's test the initial statement by considering a counterexample to disprove it:

Counterexample:
Let's consider the function f(n) = n^2.

o(f) consists of functions that grow strictly slower than f. For example, g(n) = n is in o(f) since for any constant c, when n ≥ 1/c, n < c * n^2.

O(f) consists of functions that upper bound f. In this case, let's take h(n) = n^3. It is in O(f) since for c = 1 and n ≥ 1, n^3 ≤ c * n^2.

Θ(f) consists of functions that both upper and lower bound f. In this case, let's take k(n) = n^2. It is in Θ(f) since for c1 = c2 = 1 and n0 = 1, c1 * n^2 ≤ n^2 ≤ c2 * n^2.

Now, let's calculate the set difference O(f) - Θ(f):

O(f) - Θ(f) = {h(n) = n^3} - {k(n) = n^2} = {n^3} - {n^2} = {n^3}

By comparing o(f) and O(f) - Θ(f), we can see that they are not equal. Therefore, we have found a counterexample to disprove the initial statement.

Hence, the statement "For every function f from nonnegative integers into nonnegative reals, o(f) = O(f) - Θ(f)" is false, as demonstrated by the counterexample.