 Solution

Let $I(\kappa )$, $W(\kappa )$, and $M(\kappa )$ mean respectively that $\kappa$ is inaccessible, weakly compact, and measurable. Define a fourth cardinal propertyTake ${\Phi }_0=I$, ${\Phi }_1=W$, and ${\Phi }_2=\Theta$.

If inaccessible and weakly compact cardinals exist, Question 1a shows that an inaccessible lies below every weakly compact cardinal. Thus $I{<}_{1}W$. If $W\mathbf{C}$ and $\Theta \mathbf{C}$ both hold, then $\Theta$ is exactly measurability. Every measurable cardinal is weakly compact, and the usual ultrapower reflection theorem gives weakly compact cardinals below every measurable cardinal. Therefore $W{<}_1\Theta$.

Now assume the consistency of ZFC with an inaccessible cardinal but no weakly compact cardinal. In such a model $\Theta (\kappa )$ is exactly $I(\kappa )$, soConsequently $I/{<}_1\Theta$. This is an explicit nontransitivity of the least-occurrence order on cardinal properties, even though $I{<}_{1}W$ and $W{<}_1\Theta$.