Conjecture about the number of distinct irreducible factors in a semigroup?

This might be something trivial I missed but I was wondering.

intro

Let $w(n)$ be the number of distinct prime factors of the positive integer $n$.

We know that $w(n)$ grows slowly.

It seems logical to me that

$$ \max(i<n,w(i))=A$$ $$\to w(n) < A+2$$

The simpler analogue conjecture for the count of total prime factors can be argued with numbers of the form $2^a 3^b$ taking the lead.

But this case seems a bit harder.

However if we consider that $x p$ must occur before $x p q$ then it is suddenly easy to see.

But lets generalize.

Consider the CIMAS : Cancellative Infinite Monotone Abelian Semigroups.

Let me explain that ;

1. Its a cancellative Semigroup.

2. Its an infinite Semigroup.

3. Its commutative or Abelian.

4. it is monotone : for nonzero $A,B$ we have $A*B > A*(B-1)$ and $A*B > A$

5. The elements can be ordered by size with a "norm" $p(x) = n$.

6. It is a countable set.

Conjecture 1 :

For every CIMAS we have

$$ \max(i<n,w(i))=A$$ $$\to w(n) < A+2$$

Notice we might not have unique factorization so in this context we talk about irreducibles.

Conjecture 2 :

In particular for the CIMAS example defined as

$A,B$ positive integers.

$$A*B = AB + [A g][B g]$$

Where $g$ is the golden mean, $*$ is the semigroup operator, AB is usual product of $A,B$ and $[]$ is rounding downward.

(And yes this is associative indeed and I assume no unique factorization here)

How to prove that ?