Question about Theorem 2 in CRYSTALS-Kyber Paper

I have some questions about the Kyber paper, especially about Theorem 2 on page 6, which I would like to ask here. First of all I quote the following theorem from the paper and ask my questions afterwards.

Theorem 2. For any adversary A, there exists an adversary B such that $Adv_{Kyber.CPA'}^{cpa}(A) \leq 2 \cdot Adv_{k+1,k,\eta}^{mlwe}(B)$.

1. My first question is about the definition of $Adv_{Kyber.CPA'}^{cpa}(A)$, how exactly is this defined here? If I understand CPA correctly, the idea in game $G_0$ is that the adversary picks two plaintext messages, transmits them to the challenger, and the latter randomly encrypts one message. The adversary then has to guess which message the challanger has chosen (the bit). In this setup, all variables as defined in algorithms 1-3 (in the paper) remain the same? But how exactly is $Adv_{Kyber.CPA'}^{cpa}(A)$ actually defined, what are the parameters?

2. In the change from game 0 to game 1, the $\mathbf{t}$, which was not uniformly distributed before, is replaced in game 1 with $\mathbf{t}' = \mathbf{As} + \mathbf{e}$ which is now uniformly randomly distributed. Now it is claimed that there is an adversary $B$ for which holds: $|Pr(b=b' \text{ in game } G_0) - Pr(b=b' \text{ in game } G_1)| \leq Adv_{k,k,\eta}^{mlwe}(B) \leq Adv_{k+1,k,\eta}^{mlwe}(B)$. I have some questions about this and the equation.

• First, how would such an adversary $B$ look like? If I understand correctly, it would have to try to distinguish between the games 0 and 1? The difference in the games consists here only in the $\mathbf{t}$? Just an idea, could we then perhaps define the adversary $B$ to take as input a $\mathbf{t}$ and otherwise proceed as in the encryption, so that the values $\mathbf{u}$ and $v$ then follow the distribution of $\beta$ if we are in game 0 ($\mathbf{t} = \mathbf{t}$) and otherwise follow the uniformly random distribution if we are in game 1 ($\mathbf{t} = \mathbf{t}'$)? Now if $B$ could distinguish games 0 and 1, surely it could also distinguish the samples from each other, then MLWE would not be harder than distinguishing the games? Would this be an approach? Unfortunately, it is not clear from the paper how the adversary $B$ is specifically defined.

• How is the inequality $|Pr(b=b' \text{ in game } G_0) - Pr(b=b' \text{ in game } G_1)| \leq Adv_{k,k,\eta}^{mlwe}(B)$ to be interpreted here? Does this mean that distinguishing games 0 and 1 is harder than the advantage of the adversary in ordinary MLWE? Why would that be the case? Or is it rather that the distinction of the games is not harder than the advantage of the adversary in MLWE?

• Last, I am interested in how to say that $Adv_{k,k,\eta}^{mlwe}(B) \leq Adv_{k+1,k,\eta}^{mlwe}(B)$ holds? That would mean that the advantage of the adversary in the setting $(k,k,\eta)$ is smaller than in the setting with $(k+1,k,\eta)$, how can that be? I thought with increasing $k$ the complexity of MLWE increases and thus the security increases, an attacker should find it harder to extract an advantage? This question goes back to the right of the previous paragraph on interpreting the inequality.

I hope that even though the questions are very technical, they are understandable so far. I would be very happy to receive helpful comments and answers!