To prove the theorem we need to appeal to a representation theorem for probability functions on *L* satisfying Ex. First we introduce some notation.

For the language

*L* as above a

*state description* for

\(a_1,\ldots , a_n\) is a sentence of

*L* of the form

$$\begin{aligned} \bigwedge _{i,j \le n} R(a_i,a_j)^{\epsilon _{i,j}} \end{aligned}$$

where the

\(\epsilon _{i,j} \in \{0,1\}\) and

\(R(a_i,a_j)^1 =R(a_i,a_j), R(a_i,a_j)^0=\lnot R(a_i,a_j)\). By a theorem of Gaifman, see [

1], or [

10, Chap. 7], a probability function on

*SL* is determined by its values on the state descriptions.

Let

\(D=(d_{i,j})\) be an

\(N \times N\) \(\{0,1\}\)-matrix. Define a probability function

\(w^D\) on

*SL* by setting

$$\begin{aligned} w^D\left( \bigwedge _{i,j \le n} R(a_i,a_j)^{\epsilon _{i,j}} \right) \end{aligned}$$

to be the probability of (uniformly) randomly picking, with replacement,

\(h(1), h(2),\ldots ,h(n) \) from

\( \{1,2, \ldots , N\}\) such that for each

\(i,j \le n\),

\( d_{h(i),h(j)}= \epsilon _{i,j}\). This uniquely determines a probability function on

*SL* satisfying Ex. (For details see e.g. [

10, Chap. 7]).

Clearly convex mixtures of these \(w^D\) also satisfy Ex. Indeed by the proof of [10, Theorem 25.1] it follows that any probability function *w* satisfying Ex can be approximated arbitrarily closely on *QFSL* by such convex mixtures. More precisely:

We can extend this representation result to probability functions satisfying additionally SN as follows.

For

\(\theta \in SL\) let

\(\theta ^\lnot \) be the result of replacing each occurrence of

*R* in

\(\theta \) by

\(\lnot R\) and similarly for matrix

*D* as above let

\(D^\lnot \) be the result of replacing each occurrence of 0/1 in

*D* by 1/0 respectively. For

*w* a probability function on

*SL* set

\(w^\lnot \) to be the function on

*SL* defined by

$$\begin{aligned} w^\lnot (\theta )=w(\theta ^\lnot ). \end{aligned}$$

Then

\(w^\lnot \) satisfies Ex and the probability function

\(2^{-1}(w+ w^\lnot )\) satisfies Ex+SN. Conversely if

*w* satisfies Ex+SN then

\(w=w^\lnot \) so

$$\begin{aligned} w = 2^{-1}(w + w^\lnot ). \end{aligned}$$

Thus every probability function satisfying Ex+SN is of the form

\(2^{-1}(v + v^\lnot )\) for some probability function

*v* satisfying Ex and conversely every such probability function satisfies Ex+SN.

Notice that if

$$\begin{aligned} w = \sum _D \lambda _D w^D \end{aligned}$$

then

$$\begin{aligned} w^\lnot = \sum _D \lambda _D w^{D^\lnot } \end{aligned}$$

and

$$\begin{aligned} 2^{-1}(w + w^\lnot ) = \sum _D \lambda _D 2^{-1}(w^D + w^{D^\lnot }). \end{aligned}$$

In particular then by Lemma

2,

Let

*w* be a probability function on

*SL* satisfying Ex and for a

\(2 \times 2\) \(\{0,1\}\)-matrix

$$\begin{aligned} E = \left[ \begin{array}{ll} e_{11} &{} e_{12} \\ e_{21} &{} e_{22} \end{array} \right] \end{aligned}$$

let

$$\begin{aligned} |E|_w = w(R(a_1,a_3)^{e_{11}} \wedge R(a_1,a_4)^{e_{12}} \wedge R(a_2,a_3)^{e_{21}} \wedge R(a_2,a_4)^{e_{22}}). \end{aligned}$$

We will omit the subscript

*w* if it is clear from the context. Notice that when

\(D=(d_{i,j})\) is an

\(N \times N\) \(\{0,1\}\)-matrix, then for

*E* as above we have

$$\begin{aligned} |E|_{w^D} = N^{-4} \sum _{i,j,r,s} d_{i,r}^{e_{11}} d_{i,s}^{e_{12}}d_{j,r}^{e_{21}} d_{j,s}^{e_{22}}, \end{aligned}$$

(4)

where

\(x^1=x, x^0= 1-x\). We will write

\(|E|_D\) in place of

\(|E|_{w^D}\).

A useful observation is that for any probability function

*w* satisfying Ex, |

*E*| is invariant under permuting rows and permuting columns so for example

$$\begin{aligned} \left| \begin{array}{ll} 1&{}0 \\ 1&{} 0 \end{array} \right| = \left| \begin{array}{ll} 0&{}1 \\ 0&{} 1 \end{array} \right| , \left| \begin{array}{ll} 1&{}1 \\ 0&{} 0 \end{array} \right| = \left| \begin{array}{ll} 0&{}0 \\ 1&{} 1 \end{array} \right| , \left| \begin{array}{ll} 1&{}0 \\ 0&{} 1 \end{array} \right| = \left| \begin{array}{ll} 0&{}1 \\ 1&{} 0 \end{array} \right| , \end{aligned}$$

$$\begin{aligned} \left| \begin{array}{ll} 1&{}0 \\ 0&{} 0 \end{array} \right| = \left| \begin{array}{ll} 0&{}1 \\ 0&{} 0 \end{array} \right| = \left| \begin{array}{ll} 0&{}0 \\ 0&{} 1 \end{array} \right| = \left| \begin{array}{ll} 0&{}0 \\ 1&{} 0 \end{array} \right| ,\end{aligned}$$

(5)

etc. We will use this observation frequently in what follows.

Let

$$\begin{aligned} X= \left| \begin{array}{ll} 1&{}1 \\ 1&{} 1 \end{array} \right| ~ + ~ \left| \begin{array}{ll} 0 &{}0 \\ 0&{} 0\end{array} \right| , Y= \left| \begin{array}{ll} 1&{}1 \\ 1&{} 0 \end{array} \right| ~ + ~ \left| \begin{array}{ll} 0 &{}0 \\ 0&{} 1\end{array} \right| , T= \left| \begin{array}{ll} 1&{}0 \\ 1&{} 0 \end{array} \right| , U= \left| \begin{array}{ll} 1&{}0 \\ 0&{} 1 \end{array} \right| , Z= \left| \begin{array}{ll} 0&{}0 \\ 1&{} 1 \end{array} \right| . \end{aligned}$$

**Proof.** We shall prove that

\(T \ge U\), the other inequalities follow similarly. Let

\(D=(d_{i,j})\) be an

\(N \times N\) \(\{0,1\}\)-matrix and assume first that

\(w= w^D \). By the above observation,

$$\begin{aligned} T~= ~\frac{1}{2}\left( \,\left| \begin{array}{ll} 1&{}0 \\ 1&{} 0 \end{array} \right| _{D}+\left| \begin{array}{ll} 0&{}1 \\ 0&{} 1 \end{array} \right| _{D}\right) ~~~~~ U~=~ \frac{1}{2} \left( \,\left| \begin{array}{ll} 1&{}0 \\ 0&{} 1 \end{array} \right| _{D}+\left| \begin{array}{ll} 0&{}1 \\ 1&{} 0 \end{array} \right| _{D}\right) \end{aligned}$$

so

\(T \ge U\) is the inequality

$$\begin{aligned} \sum _{i,j,r,s} d_{i,r} (1-d_{i,s}) d_{j,r}(1- d_{j,s})+ \sum _{i,j,r,s} (1-d_{i,r})d_{i,s} (1-d_{j,r}) d_{j,s} ~~~~~~~~\\ \quad ~~~~~~~~~~~~~~~~~ \ge \sum _{i,j,r,s} d_{i,r}(1- d_{i,s})(1-d_{j,r})d_{j,s}+ \sum _{i,j,r,s} (1-d_{i,r}) d_{i,s} d_{j,r} (1- d_{j,s}) \end{aligned}$$

which is equivalent to the sum over

*r*,

*s* of

$$\begin{aligned} \left( \sum _i d_{i,r} (1-d_{i,s})\right) ^2 +\left( \sum _j (1-d_{j,r}) d_{j,s}\right) ^2 - \,2 \left( \sum _i d_{i,r} (1-d_{i,s})\right) \left( \sum _j (1-d_{j,r}) d_{j,s}\right) \end{aligned}$$

being nonnegative, and hence clearly true. From this it follows that the result holds for convex combinations of the

\(w^D\) and hence by Lemma

2 for general

*w* satisfying Ex.

**Proof of Theorem ** 1 **.** We start with the left hand side inequality. Let

*w* be a probability function satisfying Ex+SN. If

\(w(R(s,h) \wedge (R(s,k) \rightarrow R(f,k))\) and/or

*w*(

*R*(

*s*,

*h*)) equals 0 then (

2) holds by our convention, so assume that these values are nonzero. Consider an approximation

\(2^{-1}\sum _D \lambda _D (w^D + w^{D^\lnot })\) of

*w* for the

\(\theta \) of the form

$$\begin{aligned} R(f,h)^{e_{11}} \wedge R(f,k)^{e_{12}} \wedge R(s,h)^{e_{21}} \wedge R(s,k)^{e_{22}} \end{aligned}$$

with small

\(\epsilon \) and

Open image in new window as guaranteed by Lemma

3.

For an

\(N \times N\) \(\{0,1\}\)-matrix

\(D=(d_{i,j})\), write

*u* for

\(2^{-1}(w^D + w^{D^\lnot })\). We have

$$\begin{aligned}&u(R(f,h) \wedge R(s,h) \wedge (R(s,k) \rightarrow R(f,k))= 2^{-1}( X_D+2T_D+Y_D), \\&\, u( R(s,h) \wedge (R(s,k) \rightarrow R(f,k))= 2^{-1}( X_D+2T_D+3Y_D +2U_D), \\&\qquad \qquad u(R(f,h) \wedge R(s,h))= 2^{-1}( X_D+2T_D+2Y_D), \\&\qquad \quad u( R(s,h))= 2^{-1}( X_D+2T_D+4Y_D +2U_D+2Z_D). \end{aligned}$$

Let

\(\hat{D}\) be another (not necessarily distinct)

\(N \times N\) \(\{0,1\}\) matrix. Working with approximations of

*w* for arbitrarily small

\(\epsilon \) it can be seen that to show (

2) for

*w* it suffices to demonstrate that for any pair

\(D, \hat{D}\) we have

$$\begin{aligned}&( X_D+2T_D+Y_D)( X_{\hat{D}}+2T_{\hat{D}}+4Y_{\hat{D}} +2U_{\hat{D}}+2Z_{\hat{D}}) ~~~~~~~~~~~~~~~~ \\&\qquad +( X_{\hat{D}}+2T_{\hat{D}}+Y_{\hat{D}})( X_D+2T_D+4Y_D +2U_D+2Z_D)~ \\&\ge ( X_D+2T_D+3Y_D +2U_D)( X_{\hat{D}}+2T_{\hat{D}}+2Y_{\hat{D}}) ~~~~~~~~~~~~~~~~~~~~~ \\&\qquad + ( X_{\hat{D}}+2T_{\hat{D}}+3Y_{\hat{D}} +2U_{\hat{D}})( X_D+2T_D+2Y_D). \end{aligned}$$

This simplifies to

$$\begin{aligned} 2X_D Z_{\hat{D}} +4T_D Z_{\hat{D}} +2 Y_D Z_{\hat{D}} + 2X_{\hat{D}} Z_D +4 T_{\hat{D}} Z_D +2 Y_{\hat{D}} Z_D \ge 4Y_{\hat{D}} Y_D +2 U_D Y_{\hat{D}} +2U_{\hat{D}} Y_D \end{aligned}$$

and since by Lemma

4 we have

\(Z_D \ge U_D\),

\( Z_{\hat{D}}\ge U_{\hat{D}} \), it suffices to show that

$$\begin{aligned} (X_D + 2 T_D) Z_{\hat{D}} + (X_{\hat{D}} +2 T_{\hat{D}}) Z_D \ge 2Y_{\hat{D}} Y_D. \end{aligned}$$

(6)

We have

$$\begin{aligned} X_D+2T_D= & {} \sum _{i,j}\big [\big (\sum _{r} d_{i,r}d_{j,r}\big )^2 + \big (\sum _{s}(1- d_{i,s})(1-d_{j,s})\big )^2 \nonumber \\&+~ 2\big ( \sum _{r}d_{i,r}d_{j,r}\big )\big (\sum _{s} (1-d_{i,s})(1-d_{ijs}) \big )\big ] \nonumber \\= & {} \sum _{i,j} \big ( \sum _r d_{i,r}d_{j,r} + \sum _s (1-d_{i,s})(1-d_{j,s})\big )^2 \nonumber \\= & {} \sum _{i,j} (x_{i,j} + y_{i,j})^2, \end{aligned}$$

(7)

where

$$\begin{aligned} x_{i,j} = \sum _r d_{i,r}d_{j,r}, \quad y_{i,j} = \sum _s (1-d_{i,s})(1-d_{j,s}). \end{aligned}$$

Similarly

$$\begin{aligned} Z_D = \sum _{i,j} \big (\sum _{r,s} d_{i,r}d_{i,s}(1-d_{j,r})(1-d_{j,s})\big ) = \sum _{i,j} z_{i,j}^2 \end{aligned}$$

(8)

where

$$\begin{aligned} z_{i,j} = \sum _r d_{i,r}(1-d_{j,r}), \end{aligned}$$

and, using (

5),

$$\begin{aligned} Y_D= & {} \sum _{i,j} \big (\sum _r(1-d_{i,r})d_{j,r} \big )\big ( \sum _{s} d_{i,s}d_{j,s} + \sum _s(1-d_{i,s})(1-d_{j,s})\big ) \nonumber \\= & {} \sum _{i,j} z_{i,j}(x_{i,j}+ y_{i,j}). \end{aligned}$$

(9)

Similarly for

\(\hat{D}= (\hat{d}_{i,j})\). Writing

\(u_{i,j}\) for

\(x_{i,j}+y_{i,j}\) etc., the inequality (

6) becomes

$$\begin{aligned} \big (\sum _{i,j} u_{i,j}^2\big )\big (\sum _{i,j} \hat{z}_{i,j}^2\big ) + \big (\sum _{i,j} \hat{u}_{i,j}^2\big )\big (\sum _{i,j} z_{i,j}^2\big ) \ge 2 \big (\sum _{i,j} z_{i,j}u_{i,j} \big )\big (\sum _{i,j} \hat{z}_{i,j}\hat{u}_{i,j}\big ) \end{aligned}$$

which holds since for any particular pairs

*i*,

*j* and

*g*,

*h*,

$$\begin{aligned} u_{i,j}^2 \hat{z}_{g,h}^2 + \hat{u}_{g,h}^2 z_{i,j}^2 \ge 2 z_{i,j}u_{i,j}\hat{z}_{g,h}\hat{u}_{g,h}. \end{aligned}$$

Turning to the right hand side inequality it is enough to show that

$$\begin{aligned} w(R(f,h) \wedge R(s,h)) \ge 2^{-1}w(R(s,h)), \end{aligned}$$

equivalently

$$\begin{aligned} w(R(f,h) \wedge R(s,h)) \ge w(\lnot R(f,h) \wedge R(s,h)). \end{aligned}$$

Proceeding as above (but much simpler since it does not need to involve the

\(\hat{D}\)) it is sufficient to show that

$$\begin{aligned} X_D + 2T_D \ge 2U_D + 2Z_D, \end{aligned}$$

and indeed this holds by Lemma

4.

\(\Box \) **Proof.** Starting with the bi-implication case and proceeding as in the proof of the second inequality in Theorem

1 it is enough to show that

$$\begin{aligned} X_D + 2T_D \ge 2Y_D. \end{aligned}$$

(10)

To this end notice that

$$\begin{aligned}&\qquad \qquad \qquad X_D =\sum _{r,s} \big (\big (\sum _i d_{i,r}d_{i,s}\big )^2 + \big (\sum _i (1-d_{i,r})(1-d_{i,s})\big )^2 \big ), \\&\qquad \qquad \qquad \qquad \qquad ~~ 2T_D = 2\sum _{r,s} \big ( \sum _i d_{i,r}(1-d_{i,s})\big )^2, \\&2 Y_D = \sum _{r,s} 2 \big (\big (\sum _i d_{i,r}(1-d_{i,s})\big )\big (\sum _i (1-d_{i,r})(1-d_{i,s}) + \sum _i d_{i,r}(1- d_{i,s})\big )\big (\sum _i d_{i,r}d_{i,s}\big )\big ). \end{aligned}$$

Writing

$$\begin{aligned} A_{r,s} = \sum _i d_{i,r}d_{i,s},~~~ B_{r,s} = \sum _i (1-d_{i,r})(1-d_{i,s}),~~~ C_{r,s}= \sum _i d_{i,r}(1-d_{i,s}) \end{aligned}$$

the required inequality becomes

$$\begin{aligned} \sum _{r,s} \big ( A_{r,s}^2 + B_{r,s}^2 + 2 C_{r,s}^2 - 2A_{r,s}C_{r,s} + 2B_{r,s}C_{r,s} \big ) \ge 0, \end{aligned}$$

which clearly holds.

The second inequality in the theorem can likewise be reduced to showing that \(X_D \ge Y_D\) and this follows from (10) and Lemma 4. \(\Box \)