Diagrammatic Proofs
3.3 Transformation Rules
In order to formalize Euclid’s method of superposition, we need to be able to use diagrams to model isometries: translations, rotations, and reflections. To do this, we first need the notion of a subdiagram. A primitive diagram A is a subdiagram of B if A is constructible from B using only rule C4. Next, we define a diagram T to be an super transformation diagram of Ain D (via transformationt) ifA is a subdiagram of D, D is a subdiagram of T, and there exists another diagramB and a functiont:A→B such thatBis also a subdiagram ofT, andAandBare equivalent or reverse equivalent diagrams via the mapt.Tis atransformation diagramofAinDviatifT is an super transformation diagram ofA inD viat, and no proper subdiagramS of T is still a super transformation diagram ofA inD viat. IfAand B are equivalent, then it is an unreversed transformation diagram, and if they are reverse equivalent, then it is areversedtransformation diagram. Now we can incorporate symmetry transformations into our system by adding the rules in Table 8. Note that simple rotations and translations are special cases of rule S1, and reflections are a special case of rule S2.
Each of these rules, like the construction rules, always yields a finite number of consequences when applied to a single diagram. This is be- cause the unmarked diagram array that results from applying one of these rules and then erasing all markings is a subarray of the the array that is obtained by constructing a copy ofAin the appropriate spot in D using the construction rules.
The system that contains the construction rules C0–C4, the trans- formation rules S1 and S2, and the rules of inference R1–R6 is called
TABLE8 Transformation Rules Transformation Rules
S1. (glide) Given a diagramD, the subdiagramA, a dotaand a dsegl1 ending atain A, and a dotband a dsegl2ending at binD, the result of applying this rule is the diagram array of all unreversed transformation diagrams ofA in D such that t(a) =bandt(l1) lies along the same dline asl2, on the same side ofbas l2.
S2. (reflected glide) Given a diagram D, the subdiagram A, a dotaand a dsegl1 ending atain A, and a dotband a dseg l2 ending at b in D, the result of applying this rule is the diagram array of all reversed transformation diagrams ofA inDsuch thatt(a) =bandt(l1) lies along the same dline as l2, on the same side ofbas l2.
FG(for “Formal Geometry”).
As an example of how these transformation rules work, consider the diagram found in Figure 11. It is a logical consequence of this diagram thatEF is congruent toBC, and we should therefore be able to mark it with three slash marks. This is one particular case of the rule of inference SAS, which we have already encountered as Euclid’s Proposition 4:
SAS. If a diagram contains two triangles, such that two sides of one triangle and the included di-angle are marked the same as two sides and the included di-angle of the other triangle, then the remaining sides of the triangles can be marked with the same new marker, and each of the remaining di-angles of the first triangle can be marked the same as the corresponding di-angle of the other.
InFG, SAS is a derived rule; it can be derived from our symmetry transformations along with CA and CS. The proof is essentially identi- cal to Euclid’s proof, with a lot of tedious extra cases showing all of the ways that the triangles could possibly intersect. The idea is to move the two triangles together using the symmetry transformations and to then check that they must be completely superimposed.
InFG, the proof of this case of SAS has two steps. The first step is to apply rule S1 to the diagram in Figure 11, moving triangleABC so thatA′ (= t(A)) coincides withD, and so that the imageA′B′ ofAB lies alongDE. The possible cases that result are shown in the diagram arrays in Figures 12 and 13. For the sake of readability, many of the
A
B
C
E
D
F
FIGURE11 The hypothesis diagram for one case of SAS.
markings have been left off these diagrams, although all markings that are later needed have been left. Also, properly speaking, these figures are only some of the cases that are given by rule S1, because any part of A′B′C′ that lies outside of DEF can intersectABC in any one of a number of ways. But the diagrams do show all of possible cases in whichA′B′C′ doesn’t intersectABC.
The second step is to remove all of the extra cases using the rules of inference CA and CS. All of the diagrams shown in Figure 12 except for the very first one can be eliminated by applying CS to A′B′ and DE. In Figure 13, all of the diagrams in the first four rows and the first two diagrams in the fifth row are eliminated in the same way. The rest of the diagrams can be eliminated by usingCA, except for the last two diagrams, which can also be eliminated by using CS. The cases that weren’t shown in Figures 12 and 13, in which A′B′C′ intersectsABC, can also all be eliminated using CS and CA. Thus, we have shown SAS for one particular case, in which the two original triangles don’t intersect and have the same orientation. The proof for the other cases is similar.