Agreement on overlap is insufficient for specialization compatibility

Each of $p,q,q^{\mathrm{\prime }}$ has one-node domain, so there is no distinct comparable pair within its domain and F1 makes it a condition. Since $x{<}_{T}y$, the nodes are distinct; each intersection of the domain of $p$ with that of $q$ or $q^{\mathrm{\prime }}$ is empty, so the functions agree on overlap. But $(p\cup q)(x)=0=(p\cup q)(y)$ violates the required inequality on $x{<}_{T}y$. Thus $p\cup q$ is not a condition and F1 makes $p,q$ incompatible.

The function $u=p\cup q^{\mathrm{\prime }}=\{(x,0),(y,1)\}$ has finite domain $\{x,y\}$ and its only unordered pair of distinct nodes has labels $0\mathrm{\ne }1$. Therefore it is a specializing condition and $u\supseteq p,q^{\mathrm{\prime }}$, so $u\le p,q^{\mathrm{\prime }}$. This is the explicit common bound verifying compatibility.