Are $\mathfrak a$ and $\mathfrak d$ your cup of tea? Revisited

This is a revised version (of late 2020) of [Sh:700], which is arXiv:math/0012170 . First point is noting that the proof of Theorem 4.3 in [Sh:700], which says that the proof giving the consistency $\mathfrak{b} = \mathfrak{d} = \mathfrak{u} < \mathfrak{a}$ also gives $\mathfrak{s} = \mathfrak{d}$. The proof uses a measurable cardinal and a c.c.c. forcing so it gives large $\mathfrak{d}$ and assumes a large cardinal. Second point is adding to the results of \S2,\S3 which say that (in \S3 with no large cardinals) we can force ${\aleph_1} < \mathfrak{b} = \mathfrak{d} < \mathfrak{a}$. We like to have ${\aleph_1} < \mathfrak{s} \le \mathfrak{b} = \mathfrak{d} < \mathfrak{a}$. For this we allow in \S2,\S3 the sets $K_t$ to be uncountable; this requires non-essential changes. In particular, we replace usually ${\aleph_0}, {\aleph_1}$ by $\sigma , \partial$. Naturally we can deal with $\mathfrak{i}$ and similar invariants. Third we proofread the work again. To get $\mathfrak{s}$ we could have retained the countability of the member of the $I_t$-s but the parameters would change with $A \in I_t$, well for a cofinal set of them; but the present seems simpler. We intend to continue in [Sh:F2009].