Ancient Indian Logic and Analogy

Abstract

B.K. Matilal, and earlier J.F. Staal, have suggested a reading of the ‘Nyāya five limb schema’ (also sometimes referred to as the Indian Schema or Hindu Syllogism) from Gotama’s Nyāya-Sūtra in terms of a binary occurrence relation. In this paper we provide a rational justification of a version of this reading as Analogical Reasoning within the framework of Polyadic Pure Inductive Logic.

J.B. Paris—Supported by a UK Engineering and Physical Sciences Research Council (EPSRC) Research Grant.

A. Vencovská—Supported by a UK Engineering and Physical Sciences Research Council Research Grant.

Appendix

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:

Lemma 2

For a probability function w on SL satisfying Ex and $\theta _1, \ldots , \theta _m \in QFSL$ and $\epsilon >0$ there is an and $\lambda _D \ge 0$ for each $N\times N$ $\{0,1\}$-matrix D such that $\sum _D \lambda _D =1$ and for $j=1,\ldots , m$,

$$\begin{aligned} | w(\theta _j) - \sum _D \lambda _D w^D(\theta _j)| <\epsilon . \end{aligned}$$

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,

Lemma 3

For a probability function w on SL satisfying Ex+SN and $\theta _1, \ldots , \theta _m \in QFSL$ and $\epsilon >0$ there is an and $\lambda _D \ge 0$ for each $N\times N$ $\{0,1\}$-matrix D such that $\sum _D \lambda _D =1$ and for $j=1,\ldots , m$,

$$\begin{aligned} | w(\theta _j) - 2^{-1}\sum _D \lambda _D (w^D(\theta _j) + w^{D^\lnot }(\theta _j))| <\epsilon . \end{aligned}$$

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

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

Lemma 4

For any probability function w satisfying Ex we have $T,Z \ge U$ and $X \ge 2Z, 2T$.

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 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. $\Box$

Theorem 5

Let w be a probability function on SL satisfying Ex+SN. Let h, k, s, f be distinct constants from amongst the $a_1, a_2, a_3, \ldots$.

Then

$$\begin{aligned}w(R(f,h)\,|\, R(s,h) \wedge (R(s,k) \leftarrow \!\!\!\!\rightarrow R(f,k))) \ge 1/2. \end{aligned}$$

$$\begin{aligned}w(R(f,h)\,|\, R(s,h) \wedge (R(s,k) \wedge R(f,k))) \ge 1/2. \end{aligned}$$

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. $\Box$