Notquitetrivial

Bx Examples Repository

Title: notQuiteTrivial

Version: 0.1

Type: Precise

Overview

Invented to illustrate a point about Least Change: almost the simplest case where there can be any question about what the best consistency restoration is.

Models

Let $B = \{true, false\}$ and let $$M = B$$ , $N = B \times B$ .

Consistency

$m\in M$ is consistent with $n = (s,t) \in N$ iff $$m = s$$ .

Consistency Restoration

Forward

An obvious choice is

$\overrightarrow{R}(m,(s,t)) = (m,t)$

Backward

No choice if we wish to be correct and hippocratic.

$\overleftarrow{R}(m,(s,t)) = s$

Properties

Correct, hippocratic

Variants

$\overrightarrow{R}(m,(s,t)) = (s,t)$ if $$m=s$$ , otherwise $(m,\neg t)$ .

Discussion

Is the version of $\overrightarrow{R}$ presented in the main body, or in the Variants section, better from a Least Change point of view? Most people's intuition probably favours the version in the main body: why would we modify the second component of $n \in N$ if we did not have to? But this depends on how well the presentation of this example matches the real-world situation it comes from. Suppose, for example, that in $$n = (s,t)$$ , $$s$$ records the presence or absence of a pebble - or a chocolate - in one pot, while $$t$$ records the presence or absence of a pebble or chocolate in another pot. The model space permits items to be destroyed, or created, but we might regard destroying or creating an item as a bigger change than moving one existing item from one pot to the other. If so, we might prefer the variant. The point is that even in so simple a case, the idea of "the" least change consistency restoration may not be quite as straightforward as one might assume.

References

None yet

Author(s)

Perdita Stevens

Reviewer(s)