Bx Examples Repository

Title: NonSimplyMatching

Version: 0.1

Type: Precise


This artificial example shows a bx that is matching but not simply matching.


Each model is a boolean: $M = N = \{0,1\}$


$R(m,n)$ iff $mn = 0$; that is

R 0 1
0 T T
1 T F

Consistency Restoration

There's only one choice if $R$ is to be correct and hippocratic. Writing Rf, Rb for forward and backward consistency restoration functions:

Rf 0 1
0 0 1
1 0 0
Rb 0 1
0 0 0
1 1 0


  • correct
  • hippocratic
  • matching (by bijection $f : 0 \mapsto 1, f : 1 \mapsto 0$)

but not

  • simply matching (see discussion)
  • undoable.


This is Example 8 from Stevens' paper referred to below. In terms of the equivalences $\sim_F$ and $\sim_B$ on each of $M$, $N$ defined there:

Each element of $M$ forms an equivalence class under each equivalence, and dually for $N$. Therefore there is only one choice of transversal, and we identify the elements with the equivalence classes.

Considered as a subset of $M_F \times M_B$, M is the diagonal subset {(0,0),(1,1)} where (0,0) represents 0 and (1,1) represents 1. Similarly for $N$. That is, the coordinate grids for $M$ and $N$ are both, identically:

1 1
0 0
0 1

We have here the simplest possible example in which there are two distinct elements of $M_F$ (0 and 1) compatible with one element of $N_B$ (0), and only one of those $M_F$ elements (0) is also compatible with a second, distinct element of$N_B$ (1). In other words, both columns of $M$'s coordinate grid are compatible with the 0 row of $N$'s, and the 0 column of $M$'s grid is also compatible with the 1 row of $N$'s.


author = {Perdita Stevens},
title = {Observations relating to the equivalences induced on model
sets by bidirectional transformations},
journal = {ECEASST},
volume = {49},
year = {2012},
ee = {},
bibsource = {DBLP,}


Perdita Stevens



Unless otherwise stated, the content of this page is licensed under GNU Free Documentation License.