Question: Using common notation, $\omega(n)$ is the number of distinct prime factors on $n$. Similiarly, $\Omega(n)$ is the number of prime factors of $n$, not necessarily distinct: $120=2^{3}\cdot 3 \cdot 5$ , therefore $\omega(120)=3$ , $\Omega(120) = 5$.

If we let $W\_k(N)$ be the count of numbers not exceeding $N$ with exactly $k$ distinct prime factors {$i | i \leq N , \omega (i) = k$}, is there any way or formulation to calculate it? Or at least a relatively coincise algorithm for it, which does not include factoring each number and checking for $\omega$ individually?

For example, one can easily find that $W\_3(100)=8$ , $W\_3(1000000)=379720$ , $W\_4(1000000)=208034$.

Please note here I am not looking for an asymptotic approximation but rather an exact count. And $N$ can be large, with order of magnitude of $10^{12}$ and over.

I have found that a solution exists for the $\Omega$ counterpart, known as [k-Almost Primes](https://mathworld.wolfram.com/AlmostPrime.html), which admits closed-form formulas, but as for $\omega$, I have found non.

Answer: 
Note the first and last givens, you are given "A or" something, and then "Not A" That immediately gives you that the other part of the not must be true, so you get B and C. Those then lead to give you D and E from the next two givens. That then leads to A or C. But once again, we know A is false, so...
In your attempt, the 3rd line is incorrect. If A is false then $A\lor (B\land C) $ (1st line) is equivalent to $B\land C.$ But if $A$ is false then $A\lor (B\lor C) $ (3rd line) is equivalent to $B\lor C.$ Also, you must incorporate some of the given premises into the argument to conclude that $C$ is true. Actually, the first premise $A\lor (B\land C) $ and the fifth premise $\neg A$ are sufficient.

Rule: If $\neg X$ then $(X\lor Y)\implies Y.$

Rule: $\neg ((\neg A)\land A).$

Applying these when $X$ is $((\neg A)\land A)$ and when $Y$ is $(\neg A)\land (B\land C),$ we have $$[\,(\neg A)\land (A\lor (B\land C))\,]\implies [\,((\neg A)\land A))\,\lor\, ((\neg A)\land (B\land C))\,]\implies $$ $$\implies [\,(\neg A)\land (B\land C)\,]\implies$$ $$\implies [\,(B\land C)\,]\implies C$$