
## Yields The Refinement:

(F lllr G)(E (flll, llll,rfll,rlll,rrll) G)(D errs E)(D rrll G) (D (rbrr,rllr,rlrr,rrrr) F)(C llrl G)(C (lrrl,rllr) F)(E errs F) (C (rllr,rrfr,rrlr,rrrr) E)(C errs D)(B (llrr,rrrr) G) (B (blrr, llll, llrf, llrl, llrr,rfll,rlll,rlrr, rrbl,rrll,rrrl,rrrr) E)
(B (rbrr,rlrr,rrfr,rrlr,rrrr) D)(B errs C)(A llrr G)(A rser F)
(A (frrr, lrrr,rrrb,rrrl,rrrr) E)(A (rllr,rlrr,rrrr) C)
(A (lfrr, llbr, llll, lllr, llrr, lrll, lrrr,rrlf,rrll,rrlr,rrrr) D)
(B (frrr, lrrl, lrrr,rrrr) F)(A errs B)
A scenario,
(F lllr G)(E flll G)(E errs F)(D rrll G)(D rrrr F)(D errs E)
(C llrl G)(C rllr F)(C rrlr E)(C errs D)(B llrr G)(B lrrl F)
(B llrl E)(B rlrr D)(B errs C)(A llrr G)(A rser F)(A rrrr E)
(A lrrr D)(A rllr C)(A errs B)
can be derived from this algebraically closed network. It is still deemed algebraically closed, even though it is not consistent with the same argument given above. Hence algebraic closure does not decide consistency for DRAf -scenarios.

On the other hand, algebraic closure with DRAfp detects all possible extensions of this network to that calculus as being inconsistent. Extending the consistent case with the relation (G rlrr C) yields three possible extensions for DRAfp scenarios, of which all are consistent. In fact, we get the three following consistent refinements.

| DRAf -relation   | refinement1   | refinement2   | refinement3   |
|------------------|---------------|---------------|---------------|
| (G rrll A)       | (G rrll- A)   | (G rrll- A)   | (G rrll- A)   |
| (G llrr D)       | (G llrr- D)   | (G llrr+ D)   | (G llrrP D)   |

We have found an example that shows that algebraic closure for DRAfp finds inconsistencies in constraint networks where it fails for DRAf . Does algebraic closure for DRAfp decide consistency? We can also give a negative result for this. To construct a counterexample, we begin with a configuration as in Fig. 28 We ensure with the constraints
(A errs B)(B errs C)(C errs D)(D errs E)(E errs F)(F errs A)
that the dipoles A, B, C, D, E and F form a convex hexagon. Furthermore, we ensure that the dipoles I, H and G form a continuous line by
(I efbs H)(H efbs G).

## The Constraints (F Rrrl I)(C Rrlr G)

state that the line has to lie inside the hexagon, since its start point and end point lie inside. To construct the counterexample, we just claim that the end