In Example 10.12 of the book The calculus of computation by Bradley and Manna, it is said

The theory of rationals is convex, as it is convex in a geometric sense.

How does the geometric sense of convex relate to theories being convex?

It further gives a proof by contradiction but I am unclear on what the assumptions are, what is being contradicted and how.

The theory of rationals TQ is convex, as it is convex in a geometric sense ... Each equality ui = vi of the disjunction G of is geometrically convex, but G itself is not. Consider, for example, H : x=y ∨ x=z. Let SH be the set of points satisfying H. The point (x,y,z) = (0,0,1) is included in SH , as is the point (1,0,1). However, the average of the two points, (12, 0,1) (choosing λ = 1/2), is not in SH. Indeed, choose any two points (u, u,v1) and (w,v2,w) from Sx=y and Sx=z, respectively, such that neither is in their intersection Sx=y=z (i.e., v1= u and v2= w). Then for any λ ∈ (0,1), the point

(λu +(1−λ)w, λu+(1−λ)v2, λv1 +(1−λ)w)

is neither in Sx=y nor in Sx=z. Suppose, then, that F ⇒ G : ni=1 ui = vi, but for no i ∈ {1,...,n} does F ⇒ui=vi. Then it must be the case that there are two points s1 and s2 of SF in separate subsets Sui=vi, Suj=vj, i= j, of SG. By the argument above, the points on the line segment between s1 and s2 are not in SG and thus not in SF. Then F is not geometrically convex, a contradiction. Thus, TQ is convex.


Context from the same book:

If a conjunctive formula in a convex theory implies a disjunction of equalities between variables, then it actually implies a single equality. Formally, consider a quantifier-free conjunctive Σ-formula F and a disjunction n

G: conjunction_over_i_to_n ( ui = vi ),

for variables ui and vi.

Theory T is convex if for every such F and G, if n

F ⇒ disjunction_over_i_to_n (ui = vi)


F ⇒ ui=vi for some i∈{1,...,n} .

If F implies G, then F actually implies one of the disjuncts of G.

Cross-posted from SO: https://stackoverflow.com/q/38814285/1494559 (as it lies in the intersection of the fields)


1 Answer 1


The key idea here is that for any conjunction of equations $F\equiv u_1=v_1\wedge\ldots\wedge u_k=v_k$, the set $S_F$ is convex in the geometric sense, i.e. for any two points $p,q\in S_F$, all points between $p$ and $q$ are also in $S_F$.

Now suppose that $F\Rightarrow G$, where $G$ is a disjunction of equations $x_1=y_1\vee\ldots\vee x_n=y_n$ (note that the variables $u_i,v_i,x_i,y_i$ are not assumed to be pairwise distinct). Without loss of generality, we may assume that $G$ is minimal in the sense that for each $i$, there is a point $p_i$ in $S_F$ which satisfies $x_i=y_i$, but not $x_j=y_j$ for any $j\neq i$. Any equation for which this is not the case can simply be removed from $G$ to obtain a smaller disjunction $G'$ which still satisfies $F\Rightarrow G'$. We will show that for such a minimal $G$, $n\le 1$.

Suppose not, i.e. $n>1$. Then the line segment $l$ between (say) $p_1$ and $p_2$ is contained in $S_F$ due to (geometric) convexity. Since $p_1$ and $p_2$ cannot be the same due to their definition, $l$ has infinitely many points. Therefore there must be some $i$ such that $l\cap S_{x_i=y_i}$ has infinitely many points, and so $l\subseteq S_{x_i=y_i}$; in particular, $p_1\in S_{x_i=y_i}$, implying $i=1$, and $p_2\in S_{x_i=y_i}$, implying $i=2$ for a contradiction.


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