Is there a law stating why a quark never decays weakly into a quark of the same weak-isospin ($T_3$)? So I read multiple articles on weak isospin and one of them stated that

"a quark never decays weakly into a quark of the same $T_3$".

If this is true, can someone please explain what law states this?
 A: The law is called:

"There are no flavor-changing neutral currents",

given the association of $T_3$ with electric charge—the hyper charge remaining invariant, naturally.
At tree level, it is absolute, given the celebrated flavor-diagonal couplings of the Z boson.
But at one loop, it is violated by a very-very small amount, suppressed by the GIM mechanism.
Still, the rare decay $B^0_s\to \phi \phi$ has been observed, amounting to $b\to ss\bar s$, dominated by penguin loop diagrams.

(For  leptons, this mechanism, is also the reason for the fantastically small rate anticipated for $\mu \to e \gamma$, a cleaner, non-hadronic mode of the essentially same phenomenon.)
A: The term in the standard model lagrangian which couples quarks of different flavors looks like
$$(\overline{u}_L,\overline{c}_L,\overline{t}_L)\gamma^\mu V_{CKM} \begin{pmatrix}d_L\\s_L\\b_L\end{pmatrix}W_\mu^- + \text{h.c}$$
where the row and columns are in flavor space. The up type quarks have $T_3=1/2$ and the down type quarks $T_3=-1/2$.
As you can see there are no couplings amongst the up or down type quarks with themselves.
