Q4 is solved in the positive: $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$. A point of dissatisfaction is that the most natural way to establish the result, that is, prove that $(a+b,b)$ is $(D\times D)$-generic (product-generic) over $M$, still does not go through. OOps - this fails because if $(a+b,b)$ is $(D\times D)$-generic then $a=(a+b)-b$ is Cohen-generic, contrary to the choice of $a$.

By the way, the result $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$ is a key lemma in my proof that it is true in the dominating-generic extension that every countable OD set of reals consists of OD reals, which hopefully will be arxived sooner rather than later.

Q4 is solved in the positive: $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$. A point of dissatisfaction is that the most natural way to establish the result, that is, prove that $(a+b,b)$ is $(D\times D)$-generic (product-generic) over $M$, still does not go through. OOps - this fails because if $(a+b,b)$ is $(D\times D)$-generic then $a=(a+b)-b$ is Cohen-generic, contrary to the choice of $a$.

By the way, the result $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$ is a key lemma in my proof that it is true in the dominating-generic extension that every countable OD set of reals consists of OD reals, arxived http://arxiv.org/abs/1609.01032.

Q4 is solved in the positive: $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$. A point of dissatisfaction is that the most natural way to establish the result, that is, prove that $(a+b,b)$ is $(D\times D)$-generic (product-generic) over $M$, still does not go through.

By the way, the result $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$ is a key lemma in my proof that it is true in the dominating-generic extension that every countable OD set of reals consists of OD reals, which hopefully will be arxived sooner rather than later.

Q4 is solved in the positive: $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$. A point of dissatisfaction is that the most natural way to establish the result, that is, prove that $(a+b,b)$ is $(D\times D)$-generic (product-generic) over $M$, still does not go through. OOps - this fails because if $(a+b,b)$ is $(D\times D)$-generic then $a=(a+b)-b$ is Cohen-generic, contrary to the choice of $a$.

By the way, the result $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$ is a key lemma in my proof that it is true in the dominating-generic extension that every countable OD set of reals consists of OD reals, which hopefully will be arxived sooner rather than later.