inorganic chemistry - Dissociation rates: trans-effect rule in square planar complexes

In lecture, the trans-effect was described.

A ligand $L^t$ with a higher trans-effect as $L$ (cis to $L^t$) leads to a faster dissociation of ligand $L^d$ (trans to $L^t$). I would expect that the attacking ligand $B$ is needed to make a stronger bond as the leaving ligand $L^d$ did.

In order to see if these assumptions are always valid in planar quadratic systems I've set up following problem according to the given trans-effect order:

$\ce{CO}$ has a stronger trans-effect as $\ce{CH3-}$ and therefore the ligand trans to $\ce{CO}$ will be exchanged faster as trans to the $\ce{CH3-}$. But the leaving group $\ce{I-}$ would also have a higher trans-effect as $\ce{Cl-}$ so I wouldn't expect a substitution in b). Are these statements correct?