  The arbitrary choice of an observable Q defines new observables P1, P2, P3 which, in turn, select rays in H3. So, to impose that observables P1, P2, P3 all have values means to assign numbers to rays in H3, and VC1, in particular, means that of an arbitrary triple of orthogonal rays, specified by choice of an arbitrary Q (briefly: an orthogonal triple in H3), exactly one of its rays is assigned 1, the others 0. Now, if we introduce different incompatible observables Q, Q′, Q″, … these observables select different orthogonal triples in H3. Assumption (1) of the KS theorem (which, effectively, is VD) now tells us that every one of these triples has three values, and VC1 tells us that these values must be for every triple, exactly {1, 0, 0}. What KS now shows is that, for a specific finite set of orthogonal triples in H3, an assignment of numbers {1, 0, 0} to every one of them (matching in common rays) is impossible.  Further reflection yields that while H3 is complex, it is in fact enough to consider a real three-dimensional Hilbert space R3. For we can show that if an assignment of values according to VC1 is possible on H3, then it is possible on R3.  Contrapositively, if the assignment is impossible on R3, then it is impossible on H3. So we can fulfill the conditions necessary to get the KS proof started and at the same time reduce the problem to one on R3.  Now, the equivalent in R3 of an arbitrary orthogonal triple in H3, is, again, an arbitrary triple of orthogonal rays (briefly: an orthogonal triple in R3). So, if KS want to show that, for a specific set of n orthogonal triples in H3 (where n is a natural number), an assignment of numbers {1, 0, 0} to every one of them is impossible, it is enough for them to show that, for a specific set of n orthogonal triples in R3, an assignment of numbers {1, 0, 0} to every one of them is impossible. And this is exactly what they do.