
## 3.9 Strong Composition

We are now going to prove that DRAfp has a strong composition. The following lemma will be crucial; note that it does not hold for DRAf .

Lemma 51. Let R be a DRAfp base relation. For DRAfp *base relations* R
not involving parallelism or anti-parallelism, betweenness and equality among
{sA, eA, SA,B}
21 for given dipoles A R B are independent of the choice of A and B, hence uniquely determined by R *alone.*
Proof. Let R = r1r2r3r4r5, where r5 ∈ {+, −} even if r5 this is omitted in the standard notation. Note that the assumption r5 ∈ {+, −} implies that SA,B
is defined. If r3 ∈ {b, s, i, e, f}, eA 6= sA = SA,B, hence there is no betweenness. Analogously, sA 6= eA = SA,B if r4 ∈ {*b, s, i, e, f*}. The remaining possibilities for r3r4r5 are:
1. ll+, rr-: in these cases, eA is between sA and SA,B; 2. ll-, rr+: in these cases, sA is between eA and SA,B;
3. rl-, lr+: in these cases, SA,B is between sA and eA.

Note that cases 1 and 2 cannot be distinguished in DRAf .

![0_image_0.png](0_image_0.png)

Corollary 52. Let R be a DRAfp base relation not involving parallelism or anti-parallelism. Let A R B and A0 R B0*. Then, the map* {sA 7→ sA0 ; eA 7→
eA0 ; SA,B 7→ SA0,B0} *preserves betweenness and equality.*
Lemma 53. Let R be a DRAfp base relation not involving parallelism or antiparallelism. Given dipoles A R C and A0 R C0 and points pA, pA0 , pC and pC0 21Please remember that sA = st(dpA) and eA = ed(dpA).