There are two random variables: $X$ and $Y$, which are independent. $X$ is drawn from a continuous cdf $G(x)$ with support $[a,b)$ and $Y$ is drawn from a cdf that consists of continuous part $F(y)$ with support $[a,b)$ and atom $k$ at $b$. $G$ and $F$ are not the same.
I wish to find the distribution of minimum of these two random variables.
Attempt: I guess a starting point is $\Pr[min(X,Y)\leq z] = 1 - \Pr[X\geq z] \Pr[Y\geq z]$. But I dont know how to continue further
This is actually an answer to a comment, but is too big for a comment.
For $z\geq b$ it is evident that $P(\min(X,Y)\leq z)=1$, and letting $z<b$ we have: $$\Pr(\min(X,Y)\leq z)=$$$$\Pr(\min(X,Y)\leq z\mid Y=b)\Pr(Y=b)+\Pr(\min(X,Y)\leq z\mid Y<b)\Pr(Y<b)=$$$$\Pr(X\leq z)k+\Pr(\min(X,Y)\leq z\mid Y<b)(1-k)=$$$$G(z)k+[1-\Pr(\min(X,Y)>z\mid Y<b)](1-k)=$$$$G(z)k+[1-\Pr(X>z\mid Y<b)\Pr(Y>z\mid Y<b)](1-k)=$$$$G(z)k+[1-\Pr(X>z)\Pr(Y>z\mid Y<b)](1-k)$$
Now note that $\Pr(Y>z\mid Y<b)\Pr(Y<b)=\Pr(z<Y<b)$ leading to: $$\Pr(Y>z\mid Y<b)=\frac{\Pr(Y>z)-k}{1-k}\tag1$$
In your comment you take wrongly $\Pr(Y>z)$ in stead of the RHS of $(1)$.