only for a subset of $S$) and the full ones. In the three–valued setting we will adopt the precision (information) ordering of the values: ${\mathbf{u}}\leq_i {\mathbf{t}}$ and ${\mathbf{u}}\leq_i {\mathbf{f}}$ The pair $(\{\textbf{t}, \textbf{f}, \textbf{u}\}, \leq_i)$ forms a complete meet–semilattice with the meet operation $\sqcap$ assigning values in the following way: $\textbf{t} \, \sqcap \, \textbf{t} = \textbf{t}$, $\textbf{f} \, \sqcap \, \textbf{f} = \textbf{f}$ and $\textbf{u}$ in all other cases. It can naturally be extended to interpretations: given two interpretations $v$ and $v'$ on $S$, we say that $v'$ contains more information, denoted $v \leq_i v'$, iff $\forall_{s\in S}\, v(s) \leq_i v'(s)$. Similar follows for the meet operation. In case $v$ is three and $v'$ two–valued, we say that $v'$ extends $v$. This means that elements mapped originally to ${\mathbf{u}}$ are now assigned either ${\mathbf{t}}$ or ${\mathbf{f}}$. The set of all two–valued interpretations extending $v$ is denoted $\lbrack v \rbrack_2$.

Let $v=\{a:{\mathbf{t}}, b:{\mathbf{t}}, c:{\mathbf{f}}, d:{\mathbf{u}})$ be a three–valued interpretation. We have two extending interpretations, $v'=\{a:{\mathbf{t}}, b:{\mathbf{t}}, c:{\mathbf{f}}, d:{\mathbf{t}})$ and $v''=\{a:{\mathbf{t}}, b:{\mathbf{t}}, c:{\mathbf{f}}, d:{\mathbf{f}})$. Clearly, it holds that $v \leq_i v'$ and $v\leq_i v''$. However, $v'$ and $v''$ are incomparable w.r.t. $\leq_i$.

Let now $w=\{a:{\mathbf{f}}, b:{\mathbf{f}}, c:{\mathbf{f}}, d:{\mathbf{t}})$ be another three–valued 