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\title{Investigating a Generalization of Probabilistic Material
  Implication and Bayesian Conditionals}

% The standard author block has changed for UAI 2023 to provide
% more space for long author lists and allow for complex affiliations
%
% All author information is authomatically removed by the class for the
% anonymous submission version of your paper, so you can already add your
% information below.
%
% Add authors
\author[1]{Michael Jahn}
\author[1]{Matthias Scheutz}
%% Add affiliations after the authors
\affil[1]{%
Human-Robot Interaction Laboratory
Department of Computer Science\\
Tufts University,
Medford, MA 02155, USA
}
  
\begin{document}
%\setcounter{page}{0}
%\input{corrections}
\maketitle

\begin{abstract}
Probabilistic ``if A then B'' rules are typically formalized as
Bayesian conditionals $\Pr(B \given A)$, as many (e.g., Pearl) have
argued that Bayesian conditionals are the correct way to think about
such rules. However, there are challenges with standard inferences
such as modus ponens and modus tollens that might make probabilistic
material implication a better candidate at times for rule-based
systems employing forward-chaining; and arguably material implication
is still suitable when information about prior or conditional
probabilities is not available at all.  We investigate a
generalization of probabilistic material implication and Bayesian
conditionals that combines the advantages of both formalisms in a
systematic way and prove basic properties of the generalized rule, in
particular, for inference chains in graphs.
\end{abstract}

\section{Introduction}
\label{sec:intro}

Natural language ``if-then'' rules cover a wide range of logically
distinct conditions such as {\em analytical truths} (``if x is an even
number, x is divisible by two''), {\em conceptual relationships} (``if
x is a human, then x is mortal''), {\em inductive inferences} (``if
the sprinkler is on, the grass will become wet''), {\em abductive
  inferences} (``if the light switch is on but the light is off, the
light bulb is broken''), {\em normative constraints} (``if the traffic
light is red, you are not allowed to drive''), {\em hypotheticals}
(``if I were to run fast, I would be out of breath''), and {\em
  counterfactuals} (``if the driver had been able to brake in time,
they would not have killed the deer'') 
  (see also \cite{ciardelli,crupi21,duboisprade90,nilsson93,greene08}).

Treating conditionals as material implications, however, has long been
seen to be problematic (e.g., \cite{adams65}) and the widely accepted
solution is to view such rules as {\em probabilistic} and best modeled
by the Bayesian conditional $\Pr(B \given A)$ \cite[e.g.,][]{pearl88}.
At the same time, Bayesian conditionals are not appropriate for
handling (indicative) conditionals \cite[e.g., see][for a
  discussion]{khoomandelkern18}, and cannot be used for inference
(e.g., using {\em modus tollens}) if the probability of the prior is
zero: $P(A)=0$.

Fortunately, there are ways to combine probabilistic logical and
Bayesian inference in a way that views probabilistic material
implication and Bayesian conditionals as two limit points of a
``rule-continuum'', $A \gimpl B$, expressed by a real-valued parameter
$\gamma\in [0,1]$ (with $\gamma=1$ for material implication $A \impl B$, and
$\gamma=0$ for the Bayesian conditional $A|B$). 

The goal of this paper is to investigate the formal properties of the
``generalized rule'' $\gimpl$ and to develop a framework that can be
used for reasoning with it.  We start with a brief motivation for the
generalization and introduce its mathematical form.  Then we start to
investigate its various properties: functional relationships among
various probabilities comparing material implication, Bayesian
conditional, and the generalized rule, as well as their bounds.  Then we
consider implication chains and prove a result on the bounds
for generalized implication chains.  We conclude with a brief discussion
and summary of our findings.

\section{Motivation and Related Work}

Efforts to combine probability and logic go back at least to Leibniz,
Jakob Bernoulli, and Boole (e.g., see \cite{hailperin96}, and also
\cite{hailperin84} for a formal framework for ``probability
models''). Mapping logical operators onto set-theoretic operators
%$\land$ onto ``$\cap$'', ``$\lor$'' onto ``$\cup$'', and ``$\lnot$'' onto ``$^{-}$'', 
in the usual way, we get set-theoretic, and hence, probabilistic
interpretations of propositional sentences, e.g., from $A \impl B
\equiv \neg A \vee B$ we get the standard
\begin{equation} \label{eqn:m=1-a+x}
    \Pr(A \rightarrow B) = \Pr(\ol{A}) + \Pr(A \intersect B) .
\end{equation}

It was recognized at least as early as \cite{boole16} (see also
\cite{hailperin84}) that bounds are needed when dealing with
probabilities of events (i.e., sets representing propositional
sentences).  Using linear programming, the existence of lower and
upper bound functions for the probability of any propositional logic
formula was established in \cite{hailperin65}.  In \cite{hailperin84},
these results were extended, and probability bounds on \modusponens
and \hypotheticalsyllogism were given.

For our purposes, the most salient motivation for working with
imprecise, i.e., bounded, probabilities is that they are {\em
  unavoidable} if we want to use \modusponens with probabilistic
implication rules: given $a=\Pr(A)$ and $m=\Pr(A \impl B)$, applying
\modusponens to infer $b=\Pr(B)$ only yields bounds on $b\in[x,m]$
where $x=\Pr(A\intersect B)$.  A formal semantic method to compute the
bounds on the probability of a given sentence in the predicate
calculus was developed in \cite{nilsson86} and extended in
\cite{faginhalpern89}.  Note that we can also use the Bayesian
conditional $B\given A$ for inference (e.g., instead of material
implication) to get bounds on $B$ which were initially presented in
\cite{wagner04} (see below for details).

An abstract notion of probabilistic logical implication $A \gimpl B$
was introduced in \cite{nguyen02}, and the probability of such a
generalized implication was computed as a function of a parameter
$\gamma$, $0 \le \gamma \le 1$:

\begin{align}
\label{eqn:g_NMK02}
\Pr(A \gimpl B) = 1- \frac{\Pr(A)(1-\Pr(B \given A))}{\gamma + (1-\gamma)\Pr(A)}\\
= 1-\frac{a(1-c)}{\gamma + (1-\gamma)a}\nonumber
\end{align}

Alternatively, we can write this generalized rule as 
\begin{align}
\Pr(A \gimpl B) = \frac{\Pr(A \intersect B) + \gamma\!\cdot\!\Pr(\ol{A})}{\Pr(A) + \gamma\!\cdot\!\Pr(\ol{A})}\nonumber
\end{align}

\noindent which more clearly shows 
the form to be some sort of interpolation from
none of $\comp{A}$ to all of $\comp{A}$.

It is immediately apparent that $\gamma=0$ corresponds to the Bayesian
conditional, and $\gamma=1$ corresponds to material implication, hence
$\gamma$ is interpolating between the Bayesian conditional and
material implication.  Hence, these two interpretations of the phrase
``if A then B'' are actually different manifestations of the same
abstract concept---the two endpoints of a one-parameter family of
probabilistic logical implications.
Figure~\ref{fig:gamma-complement-A} indicates how to view (\ref{eqn:g_NMK02}) in
a Venn diagram.

\begin{figure}
\centering
\begin{tikzpicture}[scale=0.4, opacity=0.5]
\draw[help lines] (0,0) grid (10,10);
\draw (0,0) rectangle (10,10);
\filldraw[thick, color=red,  fill=red!20, draw opacity=0.5](1,2) rectangle (6,6);
\filldraw[thick, color=blue, fill=blue!20, draw opacity=0.5](2,4) rectangle (7,9);
\filldraw[ultra thick, color=black!50!green, fill=green!20, draw opacity=0.5](6,7) rectangle (9,3);
\draw[ultra thick] (2,4) rectangle (6,6);
\node (X) at (4,5){$A  \intersect  B$};
\node (A) at (2,3){$A$};
\node (B) at (6,8){$B$};
\node (Y) at (8,4){$Y$};
\end{tikzpicture}      
\caption{
$\Pr(A)=0.20$,\ $\Pr(B)=0.25$,\ $\Pr(A\intersect B)=0.08$, 
$\gamma=0.15$, so
$\Pr(Y)=\gamma\Pr(\comp{A})=(0.15)(0.80)=0.12$. 
Then
$\Pr(A \gimpl[.15] B) = \frac{0.08+0.12}{0.20+0.12} = 0.625$,\
$\Pr(A \gimpl[0] B)=\Pr(B \given A)=0.40$,\
and
$\Pr(A \gimpl[1] B) = \Pr(A \impl B)=0.88$.
}
\label{fig:gamma-complement-A}
\end{figure}

For the rest of the paper, we will use the notational conventions in Definition~\ref{def:basic_notation}.
\begin{definition}[Notation for basic probability variables]
\label{def:basic_notation}
Fix $A,B$,$\gamma$ and let $X = A \intersect B$. Then
$a = \Pr(A)$,                         
$b = \Pr(B)$,                         
$x = \Pr(X)$,                        
$m = \Pr(A \impl B)$ for \textit{m}aterial implication,  
$c = \Pr(B\given A)$ for Bayesian \textit{c}onditional,  
$g = \Pr(A \gimpl B)$ for \textit{g}eneralized implication. 
\end{definition}
We can use the notation from Definition~\ref{def:basic_notation} to express
the four basic regions in terms of $a,b,m,c$ as 
$\Pr(A \intersect B)                 = ac        = m+a-1$,\       
$\Pr(A \intersect \comp{B})          = a(1-c)    = 1-m$,\     
$\Pr(\comp{A} \intersect B)          = b-ac      = 1-m-a+b$,\     
$\Pr(\comp{A} \intersect \comp{B})   = 1-a-b+ac  = m-b$.        
%
\newcommand{\spaceEq}{\hspace{0.1cm}=\hspace{0.1cm}}
Note that if $a=0$ then $x=\Pr(A \intersect B)=0$, even though $c$ is undefined, and the
expressions are still correct using this interpretation of $ac$ 
(as the intersection).
But typically, we work under the assumption that $a \ne 0$, so that $c$ is defined.
Furthermore, since $a,b,x,m,c$ are probabilities, we obviously must have
$0 \le a,b,x,m,c,g \le 1$. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{The Generalized Implication} 

We start by investigating the relationships among the 
(probabilities of the) generalized
implication $A \gimpl B$, the material implication $A \impl B$, the
Bayesian conditional $B \given A$, the premise $A$ and the
intersection $A\intersect B$.

\subsection{Functional relationships among $m,c,a,x,g$} 

We first express all of our basic probabilities in terms of the
classical probabilities $m,c,a,x$:

\begin{lemma}
\label{lem:fn(mcax)}
For $m,c,a$ as in Definition~\ref{def:basic_notation},
we have the following functional relationships: 
\begin{enumerate}
\item $\dsty m   \spaceEq 1-a+ac            \spaceEq 1-a(1-c)$,  
    \label{lem:fn(mcax):m}
\item $\dsty c   \spaceEq \frac{m+a-1}{a}   \spaceEq 1 - \frac{1-m}{a}$, \quad for $a \ne 0$, 
\item $\dsty a   \spaceEq \frac{1-m}{1-c}$, \quad for $c \ne 1$.    
\end{enumerate}
Furthermore, using these we can express $x=\Pr(A \intersect B)$ in the following ways:
\begin{enumerate}[resume]
\item $\dsty x   \spaceEq ac  \spaceEq a+m-1  \spaceEq \left(\frac{1-m}{1-c}\right)c$. 
\end{enumerate}
\end{lemma}
\begin{proof}
These are all just restatements of (\ref{eqn:m=1-a+x}), or the definition of $c$.
\end{proof}

The following relationship between $m = \Pr(A \impl B)$ and $c=\Pr(B \given A)$
is an easy consequence of Lemma~\ref{lem:fn(mcax)} and the definition of conditional probability.
\begin{lemma}
\label{thm:materialvsconditional}
$\Pr(A \rightarrow B)=m \geq c=\Pr(B \given A)$, in fact,
\begin{equation*}
  m
    \begin{cases}
      = 1       & \tx{if $a = 0$}, \\
      = c = b   & \tx{if $a = 1$}, \\
      = c = 1   & \tx{if $0 < a < 1$ and $a = x$}, \\
      > c       & \tx{if $0 < a < 1$ and $a \neq x$}.
    \end{cases}
\end{equation*}
\end{lemma}
\noindent
\textit{Proof.} 
For the inequality $m>c$, first note that $a \ne x \Impl c<1$. Then compute $m-c = (1-a)(1-c)>0$.
\qed

Figure~\ref{fig:mca-surfaces} shows the graphs of
the functions in Lemma~\ref{lem:fn(mcax)}.
These are, of course, the same piece of the quadratic surface of $1-m-a+ac=0$ with the axes rotated (but compare to the higher-dimensional case in Figures~\ref{fig:g(a,c)-gamma-interpolation}, \ref{fig:g(a,m)-gamma-interpolation}).
However, these different functional expressions are needed when doing iterated updates (see Figure~\ref{fig:update-example})
    and when computing bounds.

\begin{figure}
\hf
\parbox[t]{1in}{\hf \includegraphics[width=1in]{m(a,c).pdf} \hf 
                \\ 
                \hf \footnotesize$m(a,c)$ \hf }
\hf
\parbox[t]{1in}{\hf \includegraphics[width=1in]{c(a,m).pdf}  \hf
                \\ 
                \hf \footnotesize$c(a,m)$ \hf }
\hf
\parbox[t]{1in}{\hf \includegraphics[width=1in]{a(m,c).pdf}  \hf
                \\ 
                \hf \footnotesize$a(m,c)$ \hf }
\hf
\captionof{figure}{Surfaces coming from $m,c,a$}
\label{fig:mca-surfaces}
\end{figure}

The next lemma extends (and subsumes) Lemma~\ref{lem:fn(mcax)} by
expressing everything in terms of arbitrary $\gamma$, $g$ rather than
$\gamma=0,g=c$ and $\gamma=1,g=m$, as before.

\begin{lemma}
\label{lem:fn(g)}
Let $m,c,a,x,\gamma,g$ be as in Definition~\ref{def:basic_notation} and Lemma~\ref{lem:fn(mcax)}. Then
\begin{enumerate}
\item 
\label{lem:fn(g):g(mcax)}   
    $\begin{aligned}[t]
    g   
    &\spaceEq \frac{x + \gamma(1-a)}{a + \gamma(1-a)} % this line added to pre-corrections
    &&\hspace*{-1em}\spaceEq \frac{(1-\gamma)x + \gamma m}{(1-\gamma)a + \gamma\!\cdot\!1}    
    \\
    &\spaceEq \frac{ac + \gamma(1-a)}{a + \gamma(1-a)}                
    &&\hspace*{-1em}\spaceEq \frac{(a+m-1) + \gamma (1-a)}{(1-\gamma)a + \gamma\!\cdot\!1}
    \end{aligned}$
 
\item  
\label{lem:fn(g):x}
    $\dsty x        = (1-\gamma)ag + \gamma(a+g-1)$,

\item   
    $\dsty m        = (1-\gamma)ag+\gamma(a+g-1) + (1-a)$,                 
\label{lem:fn(g):m}

\item  
    $\dsty c        = \frac{(1-\gamma)ag + \gamma(a+g-1)}{a}$,               
\label{lem:fn(g):c}

\item  
    $\dsty a        = \frac{\gamma(1-g)}{\gamma(1-g) + g - c}$,             
\label{lem:fn(g):a}

\item  
    $\dsty \gamma   = \frac{a(g-c)}{(1-a)(1-g)}$,                         
\label{lem:fn(g):gamma}
\end{enumerate}
% pre-correction:    and for arbitrary $\hat{\gamma}$, with $\hat{g}=g(\hat{\gamma},x,a)$ 
    and for arbitrary $\hat{\gamma}$ (not necessarily equal to $\gamma$), 
    and letting $\hat{g}=g(\hat{\gamma},x,a)$ from part (i) above, we have
\begin{enumerate}[resume]
\item 
\label{lem:fn(g):ghat}
    $\dsty \hat{g} = \frac{[(1-{\gamma})a{g} + {\gamma}(a+{g}-1)] + \hat{\gamma}(1-a)}
                          {a+\hat{\gamma}(1-a)}$,
\end{enumerate}
assuming the variables are in the domains of the relevant formulas.
\end{lemma}
\begin{proof}
The first six come from (\ref{eqn:g_NMK02}) and straightforward arithmetic.
To get Lemma~\ref{lem:fn(g):ghat} just substitute $x({\gamma},{g},a)$ from Lemma~\ref{lem:fn(g):x}
into $\hat{g}=\hat{g}(\hat{\gamma},a,x)$ from Lemma~\ref{lem:fn(g):g(mcax)}. 
\end{proof}

Note that all four of the expressions in Lemma~\ref{lem:fn(g):g(mcax)} have
the form: an interpolation $\text{(some expression for) } x \mapsto \text{(some expression for) } m$ 
divided by an interpolation $a \mapsto 1$.  It is in this sense that $g$ interpolates from $c$ to $m$
as $\gamma$ goes from $0$ to $1$.
%
It is interesting to note that Lemma~\ref{lem:fn(g):x}, the expression
for $x(\gamma,g,a)$ is an interpolation from one expression
of $x$ ($=ac$) to another ($=a+m-1$), i.e., $x(a,c) \mapsto x(a,m)$, as $\gamma$ goes
from $0$ to $1$.  The analogous claim is also true for
Lemma~\ref{lem:fn(g):m}, \ref{lem:fn(g):c}.
%
The point of Lemma~\ref{lem:fn(g):ghat} is that the $g$ defined in
terms of one $\gamma$ can also be defined in terms of another
$\hat{\gamma}$ and its corresponding $\hat{g}$.  This is just the
general version of Lemma~\ref{lem:fn(g):g(mcax)}, \ref{lem:fn(g):m}
and \ref{lem:fn(g):c}, and will be important later when we look at
implication chains.

As above, all of these functions give a surface when graphed.  An
important point, however, is that all these expressions in
Lemma~\ref{lem:fn(g):g(mcax)} give \text{different} functions when considered as functions of the same two variables $x_1, x_2$, say.  
For example,
Figure~\ref{fig:g(a,c)-gamma-interpolation} shows the graphs
$g(\gamma,a,c)$, for various $\gamma$, and
Figure~\ref{fig:g(a,m)-gamma-interpolation} show the same series for
$g(\gamma,a,m$).
\begin{figure}
\hf
\parbox[t]{1in}{\hf \includegraphics[width=1in]{g(a,c)-gamma=0.01.pdf} \hf 
                \\ 
                \hf \footnotesize$\gamma=0.01$ \hf }
\hf
\parbox[t]{1in}{\hf \includegraphics[width=1in]{g(a,c)-gamma=0.5.pdf}  \hf
                \\ 
                \hf \footnotesize$\gamma=0.5$ \hf }
\hf
\parbox[t]{1in}{\hf \includegraphics[width=1in]{g(a,c)-gamma=1.0.pdf}  \hf
                \\ 
                \hf \footnotesize$\gamma=1.0$ \hf }
\hf
\captionof{figure}{$g(\gamma,a,c)$ for various $\gamma$}
\label{fig:g(a,c)-gamma-interpolation}
\end{figure}

\begin{figure}
\hf
\parbox[t]{1in}{\hf \includegraphics[width=1in]{g(a,m)-gamma=0.01.pdf} \hf 
                \\ 
                \hf \footnotesize$\gamma=0.01$ \hf }
\hf
\parbox[t]{1in}{\hf \includegraphics[width=1in]{g(a,m)-gamma=0.5.pdf}  \hf
                \\ 
                \hf \footnotesize$\gamma=0.5$ \hf }
\hf
\parbox[t]{1in}{\hf \includegraphics[width=1in]{g(a,m)-gamma=1.0.pdf}  \hf
                \\ 
                \hf \footnotesize$\gamma=1.0$ \hf }
\hf
\captionof{figure}{$g(\gamma,a,m)$ for various $\gamma$}
\label{fig:g(a,m)-gamma-interpolation}
\end{figure}


\subsection{Bounds on probabilities}
Henceforth, all probabilities will be \textit{imprecise}, i.e., we will typically
not know the value $p$ of a probability, but rather, will only be given
$p_0,p_1\in[0,1]$ such that $p_0 \le p \le p_1$.
The quantity $p_0$ is the \textit{lower}, and $p_1$ is the \textit{upper} probability for $p$.
Of course, there is always the need to restrict to the interval $[0,1]$
for the output of any formula giving probability bounds.
In order to simplify notation, we will build this into our notation for probability bounds.
\begin{definition}[The Lower- and Upper-bound operators]
\label{def:LU}
Let $p_0$, $p_1$ be values for
the lower and upper bounds on probability $p$, i.e., $p_0 \le p \le p_1$,
    but that $p_0$, $p_1$ are computed
from functions that do not necessarily give values in $[0,1]$.
Then
\begin{align*}
\L{p} &= \max\{ p_0, 0 \} \ , \quad   \U{p} = \min\{ p_1, 1 \}.
\end{align*}
For any probability $p$, 
we will write
$p \in \bounds{p}$, 
and we will denote the width of any probability interval $\bounds{p}$ as
$\Delta p = \U{p}-\L{p}$.
\end{definition}

We next compute bounds on one of $a,c,m,b$ given bounds on the other
variables by observing that the relevant function is monotone
increasing or decreasing (obvious from the graphs above), and then
computing the relevant endpoints to get the extreme values:

\begin{lemma}
\label{lem:partials}
Every function in 
Lemmas~\ref{lem:fn(mcax)} and \ref{lem:fn(g)} 
is of the form
$u(v^1,\ldots,v^k)$, for some $u,v^1,\ldots,v^k \in \{a,m,c,x,\gamma,\hat{\gamma},g,\hat{g} \}$. 
Fix such $u(v^1,\ldots,v^k)$ and $v\in\{ v^1,\ldots,v^k \}$ 
such that if $u(v^1,\ldots,v^k) = \hat{g}(\hat{\gamma},\hat{g},\gamma,a)$, then $v \ne a$.
Then
$\dfrac{\partial u}{\partial v}(v^1,\ldots,v^k)$ is either always positive, 
    for all $v^1,\ldots,v^k\in(0,1)$, 
or it is always negative, 
    for all $v^1,\ldots,v^k\in(0,1)$, 
assuming of course, that the partial is defined for these $v^1,\ldots,v^k$. 
\end{lemma}

We note that the one case not covered by Lemma~\ref{lem:partials}, i.e.,
   $u=\hat{g}(\hat{\gamma},\hat{g},\gamma,a)$, $v=a$ 
   (see Lemma~\ref{lem:fn(g):ghat}),
satisfies
        $
        \dfrac{\partial \hat{g}}{\partial a}(\hat{\gamma},\hat{g},\gamma,a)
        =
        \dfrac{(1-{g})({\gamma}-\hat{\gamma})}
              {a+\hat{\gamma}(1-a)} 
        $
        $<0$ \ \text{if} $\gamma<\hat{\gamma}$, \ \  
        $=0$ \ \text{if} $\hat{\gamma}=\gamma$, \ \text{ and }
        $>0$ \ \text{if} $\gamma>\hat{\gamma}
        $, 
and hence, the function $\lambda a \,\hat{g}(\hat{\gamma},\hat{g},\gamma,a)$ is decreasing if 
$\gamma<\hat{\gamma}$ and increasing if ${\gamma}>\hat{\gamma}$. 

Some specific examples of the computations in Lemma~\ref{lem:partials} are:
${\partial m}/{\partial c} = a >0$,                     
${\partial c}/{\partial a} = (1-m)/{a^2} > 0$,  
and ${\partial a}/{\partial m} = -{1}/(1-c) < 0$. 
We remark that the negative derivatives are to be expected, since they are 
a consequence of the cyclic chain rule. For example, 
  $\dfrac{\partial m}{\partial c}\dfrac{\partial c}{\partial a}\dfrac{\partial a}{\partial m} = -1$.

An easy application of Lemma~\ref{lem:partials} is the observation that the probabilities of 
generalized rules are monotonic in $\gamma$: 
\begin{lemma}
If $a,c<1$, then $\gamma_1<\gamma_2$ implies\\
$\Pr(A\gimpl[{\gamma_1}] B)<\Pr(A\gimpl[\gamma_2]B)$.
\end{lemma}
\begin{proof}
From 
$\dfrac{\partial g(\gamma,a,c)}{\partial \gamma} = \dfrac{a(1-a)(1-c)}{( a+\gamma(1-a) )^2} >0$.
\end{proof}


\subsection{Bounds on $m,c,a,g$ from partial derivatives}

An immediate consequence of Lemma~\ref{lem:partials} is that 
all these functions
      are monotonic on $(0,1)$ in each of their variables,
      with the exception of the one in Lemma~\ref{lem:fn(g):ghat}
              that has a min when $\hat{\gamma}=\gamma$.
So to compute the
bounds of any of our basic probabilities, in terms of the bounds of
other variables, we just need to evaluate at the appropriate
endpoints, i.e., the bounds of the independent variables.

\begin{lemma}
\label{lem:bounds}
Fix any function from Lemma~\ref{lem:partials}, $u(v^1,\ldots,v^k)$, 
with $u,v^1,\ldots,v^k\in\{a,m,c,x,\gamma,g\}$, 
and $v^i \in \bounds[i]{v}$.
Let $\E[i]{v} = 
\begin{cases} 
\L[i]{v} & \text{if } \partial u/\partial v^i > 0 \\ 
\U[i]{v} & \text{if } \partial u/\partial v^i < 0 
\end{cases}$
\ and \ $\F[i]{v} = 
\begin{cases} 
\L[i]{v} & \text{if } \partial u/\partial v^i < 0 \\ 
\U[i]{v} & \text{if } \partial u/\partial v^i > 0 ,
\end{cases}$
and note that $\{ \E[i]{v},\F[i]{v} \} = \{ \L[i]{v},\U[i]{v} \}$.
Then we have
the bounds
$\L{u} = u(\E[1]{v},\ldots,\E[k]{v})$ and  
$\U{u} = u(\F[1]{v},\ldots,\F[k]{v})$. 
\end{lemma}

An easy example of Lemma~\ref{lem:bounds} in action would be to use
$\partial m(a,c)/\partial a < 0$ and $\partial m(a,c)/\partial c > 0$
to get $\L{m} = m(\U{a},\L{c}) = 1-\U{a}(1-\L{c})$. 
Similarly, from $\partial a/\partial m >0$, $\partial a/\partial c<0$, we get $\U{a}=a(\L{m},\U{c}) = (1-\L{m})/(1-\U{c})$,
and from 
$\partial g/\partial \gamma >0$, $\partial g/\partial a <0$, 
and $\partial g/\partial c >0$,
we get
$\L{g}(\gamma,a,c) = g(\L{\gamma},\U{a},\L{c}) = \dfrac{\U{a}\L{c} +
  \L{\gamma}(1-\U{a})}{\U{a}+\L{\gamma}(1-\U{a})}$. 

Note that when we are computing $\L{m}$ we are treating $m$ as a dependent 
variable, and $a,c$ as independent, and when we compute the relevant partials we are holding either $a$ or $c$ constant. 
This is not the same as solving for $\L{m}$ in the formula for $\U{a}$ in order to compute $\L{m}$.
Also, if these formulas are being used to update a node in a network
as new information comes in, we would consider $\U{a},\L{c}$ to be fixed known quantities for which we want to compute the corresponding smallest consistent $m$-value, $\L{m}$.
This is clearly not the same as updating $\U{a}$ given the newly measured (and fixed) quantities $\L{m},\U{c}$.

We must also keep in mind that any updated probability is clipped 
to stay within its old bounds, as per Definition~\ref{def:LU}. 
Also, note that if there are multiple
ways to compute a bound, e.g., $\U{m'}(\L{a},\U{c})$,
$\U{m'}(\L{a'},\U{c})$, $\U{m'}(\L{a},\U{c'})$,
$\U{m'}(\L{a'},\U{c'})$, they will not always yield the same
result. We must always compute with the most recent data, since the
order that the updates come to us matters to the final answer.
The issues above are most salient in a network of nodes that each hold a probability and are being updated in real time as new information becomes available. See Figure~\ref{fig:update-example} for a example computation.

The bounds listed in Lemma~\ref{lem:bounds} can be interpreted
geometrically.  The graph of $m(a,c)$ is given in
Figure~\ref{fig:m(a,c)-bounds}, and each point on this surface gives a
consistent assignment of values to $m,c,a$.  Bounds on $a,c$
correspond to vertical planes, and bounds on $m$ to horizontal planes.
It is apparent that for even a single $m$-bound, $\L{m}$ say, the set 
$\{a,c : \L{m} \le m(a,c) \le 1 \}$ 
will be non-convex.

\begin{figure}
  \includegraphics[width=1.6in]{mca-bounds-v.pdf}  
  \includegraphics[width=1.6in]{mca-bounds-h.pdf}    
  \caption{Geometric interpretation of bounds for $m(a,c)$
    ($a,c$-bounds left, $m$-bounds right).}
  \label{fig:m(a,c)-bounds}
\end{figure}


\begin{figure*}[t]
\[
\begin{array}{ccccccc}
\text{iteration}
& \underset{1-a_1(1-c_0)}{\L{m}}              % m_0
& \quad \underset{1-a_0(1-c_1)}{\U{m}}        % m_1
& \quad \underset{1-(1-m_0)/a_0}{\L{c}}       % c_0
& \quad \underset{1-(1-m_1)/a_1}{\U{c}}       % c_1
& \quad \underset{(1-m_1)/(1-c_0)}{\L{a}}     % a_0
& \quad \underset{(1-m_0)/(1-c_1)}{\U{a}}     % a_1
\\
\hline
\hline
\text{start}    & 0           &   1   &   0   &   1   &   0   &   1   \\  
\hline
1               & \fbox{0.5}  &   1       &   \underline{0}   &   1   &   0   &   \overline{1}   \\  
\hline
2 & \underline{0.5}         & \fbox{0.8} &   \underline{0}   &   0.8 &   0.2 &   \overline{1}   \\  
\hline
3 & \underline{0.5}         & \overline{0.8}       &   {1}/{6}&   0.8 &   \fbox{0.6} &   \overline{1}   \\  
\hline
4 & \underline{0.5}         & \overline{0.8}       &   {1}/{6}&   0.75 &   \underline{0.6} &   \fbox{0.8}   \\  
\hline
\hline
\text{start}    & 0         &   1   &   0   &   1   &   0   &   1   \\  
\hline
1 & 0           & 1   & 0   &   1   &   \fbox{0.6}   &   1   \\  
\hline
2 & 0.2         & 1   & \underline{0}   &   1   &        {0.6}   &   \fbox{0.8}   \\  
\hline
3               & \fbox{0.5}    &   1   &   {1/6}   &   1   &        \underline{0.6}   &   \overline{0.8}   \\  
\hline
4 &  \underline{0.5}  & \fbox{0.8}    &   {1/6}   &   0.75   &        \underline{0.6}   &   \overline{0.8}   \\  
\hline
\hline
\text{start}    & \fbox{0.2}        &   \fbox{0.8}      &   0   &   1   &   \fbox{0.1}   &   \fbox{0.9}   \\  
\hline
1               & \underline{0.2}   &   \overline{0.8}  & \underline{0}   &  7/9  &   0.2   &   \overline{0.9}   \\  
\hline
\hline
\text{start}    & \fbox{0.3}       &  \fbox{0.8}  &   0   &   1   &   0   &   1   \\  
\hline
1               & 0.3       &  0.8  &   0   &  0.8  &   {0.2}   &   1   \\  
\hline
\end{array}
\]
\caption{
Updating bounds with new data: 
$\fbox{$q$}$ means $q$ was manually reset due to new information into to the system. 
    If $q$ was outside the old bounding box,
    $\underline{q}$ means $q$ was clipped up to the old bound $\L{q}$,
    and $\overline{q}$ means clipped down to the old $\U{q}$. 
}
\label{fig:update-example}
\end{figure*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Bounds on $m,c,g$ for fixed $a,b$}

For fixed $\bounds{a}$, $\bounds{b}$, we want to find $\bounds{c}$,
$\bounds{m}$, and $\bounds{g}$.  As a first special case, assume
$a=\L{a}=\U{a}$, $b=\L{b}=\U{b}$.  Then note that $\Pr(A \intersect B)
=x =ac$, but we do not have $c$.  Note also that $x$ is \textit{not} a
function of $a,b$, but has the following bounds: % by (\ref{bds:booleC}):
\begin{align}
\L{x} &= a+b-1 \ , & \U{x} &= \min\{ a, b \}       
\label{bds:ac(a,b)} . 
\intertext{In general, if there are nontrivial probability intervals for $a,b$, then}
\L{x} &= \L{a}+\L{b}-1 \ , & \U{x} &= \min\{\U{a},\,\U{b} \} .
\end{align}
From this we want to compute bounds on $c,m$.


Of course, we do not have a function $c(a,b)$ to minimize/maximize,
but $\frac{\partial}{\partial a}\left(\frac{a+b-1}{a}\right) \ge 0$,
$\frac{\partial}{\partial b}\left(\frac{a+b-1}{a}\right) > 0$, 
and
$\L{c}$, $\U{c}$ will occur at the bounds of $x$, so
$\bounds{c} = [ \max\{(\L{a}+\L{b}-1)/\L{a},\, 0\},\ \min\{ \U{b}/\L{a},\, 1 \} ]$.
From this $c$ we can get bounds on $m(a,c(a,b))$ as before, but there might be clipping
when recomputed using bounds gotten from $m(a,b)$ as below. Then
$\bounds{m} = [ 1-\U{a}(1-\L{c}),\ 1-\L{a}(1-\U{c})]$.

Using $m=1-a+x$, $x$ depends on $a,b$ (not $a,c$), so
$\bounds{m} = [ \L{(1-a+x)},\ \U{(1-a+x)} ]
\ne [ \underbrace{1-\U{a}+\L{(ac)}}_{\text{$a$, $ac$ not indep}},\ 1-\L{a}+\U{(ac)} ]$.
We need to minimize $m=1-a+x$, but expressed using the lower bound of $x$ from
(\ref{bds:ac(a,b)}). So we need to minimize $m=1-a+\max\{a+b-1,0\}$.  From the graph

\hf
\includegraphics[width=1in]{m(a,b)-with-ac-min.pdf}
\hf

we see $\dfrac{\partial m}{\partial a}\le0$ and $\dfrac{\partial m}{\partial b}\ge0$,
so we compute at the point $(\U{a},\L{b})$, 
$\L{m} = 1-\U{a}+\max\{\U{a}+\L{b}-1,\,0\}
       = \begin{cases}
         1-\U{a} & \text{if } \L{b} \le 1 - \U{a}    \\
         \L{b} & \text{if } \L{b}  >  1 - \U{a}   
         \end{cases}$.
And to maximize $m$ we need to use the upper bound for $x$ from (\ref{bds:ac(a,b)}). 
So we maximize $m=1-a+x = 1-a+\min\{a,b\}$. From the graph

\hf 
\includegraphics[width=1in]{m(a,b)-with-ac-max.pdf}
\hf

we see that $\dfrac{\partial m}{\partial a}\le0$ and $\dfrac{\partial m}{\partial b}\ge0$,
    so we compute at the point $(\L{a},\U{b})$, 
$\U{m} = 1-\L{a}+\min\{\L{a},\U{b}\}
              =    \begin{cases}
                    1-\L{a}+\U{b}   & \text{if } \U{b} \le \L{a}    \\
                    1               & \text{if } \U{b}  >  \L{a}     
                    \end{cases}$.

\subsection{Bounds on $b$}
\label{sec:bounds_b}

We next investigate different ways to obtain bounds on $b$.  Let
$a\in\bounds{a}$. 
If we know both $m,c$ (exactly), then we know $a$ ($=\frac{1-m}{1-c}$) exactly, so using either material
implication ($\gamma=1$) or conditional ($\gamma=0$) gives the same
bounds $b \in [ {a}c ,\ m ]$.

Now suppose we know $m$, but are not given any
information about $c$.  Hence we must compute $c\in\bounds{c} =
[({m}+\L{a}-1)/\L{a}, ({m}+\U{a}-1)/\U{a}] $.  Then $b \in [ \L{a}+m-1
  ,\ m]$, i.e.,
$b = [\L{a}\L{c} ,\ m]$, and $\Delta b  = m-\L{a}\L{c} = m-(\L{a}+m-1)
              = 1-\L{a}$.
Also note: 
$\Delta c = \dfrac{\U{a}+m-1}{\U{a}} - \dfrac{\L{a}+m-1}{\L{a}} 
    = \dfrac{\Delta a}{\L{a}\U{a}}(1-m)$.
% pre-correction: But since $a+m\ge1$, for all consistent $a$, we have
But since $a+m\ge1$, for all consistent $a$ (in particular $\L{a}$), we have
$\dfrac{\Delta a}{\L{a}\U{a}}(1-m) \le \dfrac{\Delta a}{\L{a}\U{a}}\L{a}
    = \dfrac{\Delta a}{\U{a}} = 1-\dfrac{\L{a}}{\U{a}}$. 

Next suppose we know $c$, but are not given any information about $m$.
Hence we must compute $m\in\bounds{m} = [ 1-\U{a}(1-c) ,\ 1-\L{a}(1-c)
]$.  Then
$b \in [ \L{a}c ,\ \U{m} ]$, and $\Delta b  = \U{m} - \L{a}{c} = 1-\L{a}(1-c) - \L{a}c 
              = 1-\L{a}$.
Also, $\Delta m  = 1-\L{a}(1-c) - 1 + \U{a}(1-c) = \Delta a(1-c)$.
So we see that knowing $m$ (and not $c$) gives a better $\U{b}$,
and knowing $c$ (and not $m$) gives a better $\L{b}$. 
But the $\Delta b$ is the same in both cases.

Finally, suppose we are given
$m\in\bounds{m}$ and $c\in\bounds{c}$.  Then inference with material
implication gives
$\L{b} = \L{a}{c}(\L{a},\L{m}) = \L{a}+\L{m}-1$, $\U{b} =  \U{m}$ 
which gives $\Delta b = 1-\L{a} + \Delta m$.
Bayesian inference gives $\L{b} =  \L{a}\L{c}$, 
$\U{b} =  {m}(\L{a},\U{c}) = 1-\L{a}(1-\U{c})$,
and
$\Delta b = 1-\L{a}(1- \Delta c) = 1 - \L{a} + \L{a}\Delta c$.
And if we kept the best bounds from both, we would have 
$\L{b} = \max\{ \L{a}+\L{m}-1 ,\ \ \L{a}\L{c} \}$,
$\U{b} =  \min\{ \U{m} ,\ \ 1-\L{a}(1-\U{c}) \}$.

\textit{A priori}, any of these four bounds could be optimal.
        In practice, the values of $\L{c},\U{m}$ would already be reset to the computed values 
        if the original values were less optimal than the computed ones in the 
        optimization/clipping phase.  In this case we would just have 
$b \in [ \L{a}\L{c}, \U{m} ]$.  Then 
$\Delta b = \U{m}-\L{a}\L{c}$. 
We can't simplify algebraically, but 
$\U{m}\ge m(\L{a},\U{c})$ and $\L{c}\le c(\L{a},\L{m})\le\U{c}$ give
    $\Delta b \ge m(\L{a},\U{c}) - \L{a}\L{c}   = (1-\L{a}) + \L{a}\Delta{c}$,
    $\Delta b \ge \U{m} - \L{a}c(\L{a},\L{m}) = (1-\L{a}) + \Delta{m}$.

Now we want to compute bounds on $b$ when we have a generalized implication $A \gimpl B$.
The equations in Lemma~\ref{lem:fn(mcax)} are true for any $B$ with $\Pr(B \given A)=c$,
but $b$ is itself not functionally related to any of $m,c,a$.
Consequently, we can only get bounds on $b$, even if we know all of $m,c,a,x$ precisely.
In this case, we use
$\Pr(A \intersect B) \le \Pr(B) \le \Pr(\comp{A} \union B)$, to get the well-known (tight) bounds
$x \le b \le m$, i.e.,
Using Lemma~\ref{lem:fn(g):x},~\ref{lem:fn(g):m}, we can express these bounds in terms of $g$:
\begin{align}
\L{b}   &=(1-\gamma){a}{g} + \gamma({a}+{g}-1) ,
\label{eqn:bL(a,g)}
\\
\U{b}   &= (1-\gamma){a}{g} + \gamma({a}+{g}-1) + (1-{a}) .
\label{eqn:bU(a,g)}
\end{align}
Of course, these formulas apply when we have precise probabilities for $a,g$.
    If we only have bounds for $a$ and $g$, we compute
    $\dfrac{\partial}{\partial a}\L{b}(a,g)>0$,\
$\dfrac{\partial}{\partial g}\L{b}(a,g)>0$, and
    $\dfrac{\partial}{\partial a}\U{b}(a,g)<0$,\
    $\dfrac{\partial}{\partial g}\U{b}(a,g)>0$.
Using Lemma~\ref{lem:bounds} we can compute $\L{(\L{b})}=\L{x}$, $\U{(\L{b})}=\U{x}$, $\L{(\U{b})}=\L{m}$, $\U{(\U{b})}=\U{m}$.
We will abuse notation and write $\L{b}=\L{(\L{b})}$ and $\U{b}=\U{(\U{b})}$, and discard the other two:
\begin{align}
\L{b}   &= \L{b}(\L{a},\L{g})
        = (1-\gamma)\L{a}\L{g} + \gamma(\L{a}+\L{g}-1)
\label{eqn:bL(aL,gL)}
\\
\U{b}   &= \U{b}(\L{a},\U{g}) \notag
\\      &= (1-\gamma)\L{a}\U{g} + \gamma(\L{a}+\U{g}-1) + (1-\L{a}) .
\label{eqn:bU(aL,gU)}
\end{align}
Whether $\L{b}$ refers to the function in (\ref{eqn:bL(a,g)}) or the value in (\ref{eqn:bL(aL,gL)})
will be clear from context.  Furthermore,
$\Delta{b} = (1-\L{a}) + [(1-\gamma)\L{a}+\gamma\!\cdot\!1]\Delta g$, and
if $g$ is precise we just have $\Delta{b}=1-\L{a}$, as noted above for the case
of precise $m$, $c$.

    
\section{Implication chains with generalized implication}

Implications chains (i.e., \hypotheticalsyllogism) are of particular
interest for rule-based reasoning, hence we briefly summarize the
properties of the generalized implication chains.

Suppose we have a sequence of generalized implications $B^{i}
\gimpl[\gamma^{i}] B^{i+1}$, for $i=0,1,\ldots,N\!-\!1$.  Let $A=B^0$,
and suppose we are given bounds, $a\in\bounds{a}$, on the initial
antecedent $A$.  We want to compute the bounds for the final
consequent $b^N$.  We first do this for arbitrary $\gamma$. Then we
easily derive the bounds for material implication (\modusponens,
\hypotheticalsyllogism) and Bayesian inference by setting
$\gamma=0,1$.

\begin{theorem}\label{thm:impl-chain}
Fix arbitrary $\gamma$ and suppose we are given a sequence of generalized implications 
$B^{i} \gimpl B^{i+1}$, for $i=0,1,\ldots,N\!-\!1$. 
Suppose $A=B^0$ and that we are given bounds $a\in\bounds{a}$,  
        $g^{i} \in \bounds[i]{g}$, for $i < N$.  
Let $Y^{i} = \gamma(1-{g^{i}})+{g^{i}}$ and $Z^{i} = \gamma(1-{g^{i}})$.
Then the bounds at the end of the generalized implication chain are
\begin{equation*}
\begin{aligned}
\L[N]{b}    &= \L{a}\prod_{i=0}^{N-1} \L[i]{Y} - \sum_{i=0}^{N-1}\left(\U[i]{Z}\prod_{j=i+1}^{N-1}\L[j]{Y}\right)  ,
\\
\U[N]{b}    &= 1 - \L[N-1]{Z} \,- 
            \\
            &\hspace*{2ex}
            (1-\U[N-1]{Y}) 
            \left[
                    \L{a}\prod_{i=0}^{N-2} \U[i]{Y} -
                            \sum_{i=0}^{N-2}\left(\L[i]{Z}\prod_{j=i+1}^{N-2}\U[j]{Y}\right)
             \right] .
\end{aligned}
\end{equation*}
\end{theorem}
\begin{proof}
We first note that
$\L[i]{Y}=Y^i(\L[i]{g})$,
$\U[i]{Y}=Y^i(\U[i]{g})$,
$\L[i]{Z}=Z^i(\U[i]{g})$,
$\U[i]{Z}=Z^i(\L[i]{g})$.
To prove the claim we give a straightforward induction on $n$.
For the base case $n=1$, simple substitution and algebra gives
$\L[1]{b}   = \L{a}\L[0]{Y}-\U[0]{Z}
            = (1-\gamma)\L{a}\L[0]{g} + \gamma(\L{a} + \L[0]{g} -1)$,
which is just (\ref{eqn:bL(aL,gL)}).
Similarly,
$\U[1]{b}   = 1-\L{a}(1-\U[0]{Y}) - \L[0]{Z}
            = 1-\L{a} + (1-\gamma)\L{a}\U[0]{g} + \gamma(\L{a} + \U[0]{g} -1)$,
which is just (\ref{eqn:bU(aL,gU)}).

For the inductive step, we get
$\L[n+1]{b} = \L[n]{b} \L[n]{Y} - \U[n]{Z}$
            \\$= \left[
              \L{a} \prod_{i=0}^{n-1}\L[i]{Y}
                - \sum_{i=0}^{n-1}\left( \U[i]{Z}\prod_{j=i+1}^{n-1}\L[j]{Y} \right)
              \right]
              \L[n]{Y} - \U[n]{Z}$
            \\$= \L{a} \prod_{i=0}^{n}\L[i]{Y}
                - \sum_{i=0}^{n}\left( \U[i]{Z}\prod_{j=i+1}^{n-1}\L[j]{Y} \right)\L[n]{Y} - \U[n]{Z}$
            \\$= \L{a} \prod_{i=0}^{n}\L[i]{Y}
                - \sum_{i=0}^{n}\left( \U[i]{Z}\prod_{j=i+1}^{n}\L[j]{Y} \right)\L[n]{Y}$,
as desired,
and similarly for $\U[N+1]{b}$.

Recall that $\L[i]{b}=\cdots$, $\U[i]{b}=\cdots$ 
are automatically clipped to remain in $[0,1]$, as per Definition~\ref{def:LU},
but that this clipping is lost if we combine and rearrange the RHS expressions for these quantities,
rather than referring to the LHS quantities themselves.
So the above formulas assume no clipping was necessary at intermediate stages of the recursion.
\end{proof}

Note that since $\gamma=0 \Impl Y=c, \ Z=0$, and $\gamma=1 \Impl Y=1, \ Z=1-m$, we can easily
recover the classical cases below. 

\begin{corollary}
If we are using Bayesian inference, then $\gamma=0$. So $c^{i} \in \bounds[i]{c}$, for $i < N$
are given, and the bounds at the end of the implication chain are
\begin{align*}
\L[N]{b}    &= \L{a}\prod_{i=0}^{N-1}\L[i]{c} \ , \quad
\U[N]{b}     = 1-\left(\L{a}\prod_{i=0}^{N-1}\L[i]{c}\right) \left(1-\U[N-1]{c}\right)
\end{align*}
We note that for typical data (not all $\L[i]{c}=1$) we see $\L[N]{b} \to 0$ and $\U[N]{b} \to 1$.  
\end{corollary}

\begin{corollary}
If we are using material implications in our chain, we just set $\gamma=1$ in the formulas above.
So $m^{i} \in \bounds[i]{m}$, for $i < N$, are given, 
and the bounds at the end of the implication chain are
    \begin{align*}
    \L[N]{b}    &= \L{a} - \sum_{i=0}^{N-1} \left( 1-\L[i]{m} \right) \ , \quad\quad
    %\\          &= \L{a} + \sum_{i=0}^{N-1} \L[i]{m} \ -N
    \U[N]{b}    = \U[N-1]{m} .
    \end{align*}
We note that for typical data (not all $\L[i]{m}=1$) we have $\L[N]{b} \to 0$. Also, $\U[N]{b}$ just depends on the last implication.
    %\\ $\Delta b^N = \Delta m^{N-1} + N - \L{a} - \dsty\sum_{i=0}^{N-2}\L[i]{m}$.
\end{corollary}

\begin{lemma}
Suppose we are given a sequence of generalized implications
$B^{i} \gimpl[\gamma^i] B^{i+1}$, for $i=0,1,\ldots,N\!-\!1$.
Let $A=B^0$, and suppose we are given the bounds $a\in\bounds{a}$,
and the bounds $g^{i} \in \bounds[i]{g}$, for $i < N$.
Then the bounds for the end of the generalized implication chain can be computed 
using Theorem~\ref{thm:impl-chain}.
\end{lemma}
\begin{proof}
We just use Lemma~\ref{lem:fn(g):ghat} to write all of the $\gamma^i,g^i$ pairs
in terms of a single fixed $\hat{\gamma}$ (of our choice), and the newly computed $\hat{g}^i$.
Then apply Theorem~\ref{thm:impl-chain}.
However, these bounds might not be optimal for the reasons discussed in Section~\ref{sec:bounds_b}. 
\end{proof}

\begin{lemma}
The generalized implication chain, 
$B^{i} \gimpl[\gamma] B^{i+1}$, for $i=0,1,\ldots,N\!-\!1$,
from Theorem~\ref{thm:impl-chain},
can be replaced with a single implication $A \gimpl B$, where $A=B^0$, $B=B^N$, and
the bounds $\bounds[N]{b}$ are the same, whether computed from the implication chain 
or the single implication. 
\end{lemma}
\begin{proof}
Note that everything is specified except for the bounds on $g$.
Since $\gamma$, $\L{a}$, $\L{b}=\L[N]{b}$, and $\U{b}=\U[N]{b}$ are all fixed,
equations (\ref{eqn:bL(aL,gL)}), (\ref{eqn:bU(aL,gU)}) give two linear equations
each with one unknown.
\end{proof}

Several consequences of this result are noteworthy:

  \tb{(a)}~The upper bound $\U{a}$ %$\Upper{A_0}$ 
  of the initial antecedent $A$ plays no role in the determination of the bounding interval 
  of the final consequent $B^N$. This comes as no surprise because the same is true for a single rule 
  (see (\ref{eqn:bU(aL,gU)})).
  
  \tb{(b)}~The lower bound $\L[N]{b}$ of the consequent $B^N$ is completely determined by the lower bounds of the antecedent $A$ and the rules being cascaded together; their upper bounds play no role in its determination. In fact, if the lower bound of the antecedent $A$ or any of the rules is zero, the lower bound of the consequent is zero.
  
  \tb{(c)}~In the case $\gamma=1$ (material implication) the upper bound $\U[N]{b}$ of the consequent $B^N$ is completely determined by, in fact it is equal to, the upper bound of the last rule $B^{N-1} \rightarrow B^N$; neither its lower bound nor the antecedent $A$ nor the other rules play a role in the determination of $\U[N]{b}$.
  
  \tb{(d)}~In the case $\gamma=1$ (material implication), we compute 
  $\Delta b^N = \U[N]{b}-\L[N]{b} = \Delta b^N 
    = \Delta m^{N-1} + (1-\L{a}) + \sum_{i=0}^{N-2}(1-\L[i]{m})$. 
Interestingly, this implies that $\Delta b^N \geq \Delta m^{N-1}$, %A_{N-1} \rightarrow A_N)$, 
i.e., the uncertainty interval associated with the consequent $B^N$ can never be narrower than the uncertainty interval associated with the rule at the end, viz., $B^{N - 1} \rightarrow B^N$. In fact, 
\begin{equation*}
  \Delta b^N
 \begin{cases} =
    \Delta m^{N-1}, %A_{N-1} \rightarrow A_N),
      & 
      \tx{if\ } \L[N-1]{m}=0 % \tx{if $\Lower{A_{N-1} \rightarrow A_N} = 0$ 
          \tx{\ \ or } 
      \\ &
      \ \ \ \L{a} = \L[i]{m} = 1, 
       %i \in \{0, \ldots, n-2\}; 
      \  0 \le i \le n-2;
      \\
    > \Delta m^{N-1}, %\Delta(A_{N-1} \rightarrow A_N),
      & \tx{otherwise}.
  \end{cases}
\end{equation*}

\section{Discussion}

Our investigation of the properties of the generalized rule revealed
several ways for expressing and computing bounds.  In particular, it
showed the close relationship between probabilistic material
implication and Bayesian conditional, especially when used for
inference with rules like modus ponens.  Paired with the generalized
rule, inference principles like modus ponens can generate bounds on
the consequent that depend both on the $\gamma$ value and the rule
bounds, and the theorem about implications chains shows that it is
possible to calculate inferences from any mixture of rule bounds and
$\gamma$ values (recursively in the worst case when clippings happens, 
see the discussion after Lemma~\ref{lem:bounds}).
This then enables mixed rule-based systems with different $\gamma$
values for rules, e.g., high $\gamma$ values for rules that are based
on little data and are more conceptual in nature, and low $\gamma$
values for rules for which sufficient data exists.  In general, one
possible application of the $\gamma$ parameter is to make it a
function of the amount of data from which rules have been extracted:
the more data is available, the closer the representation should go do
the normatively correct Bayesian end point.
The issue of the interpreting the generalized rule remains, and is not addressed here, 
but the starting point for that investigation would be the observation that if we consider the range of the conditional going down from  $\top$ to $A$, i.e., 
going from $\Pr(A \impl B \given \top) = \Pr(A \impl B)$  all the way to $\Pr(A \impl B \given A)$, 
we will get exactly the generalized rule.

\section{Conclusion and Future Work}

The goal of this paper was to investigate the relationship of
probabilistic material implications and Bayesian conditionals, and for
this purpose we utilized a generalization that subsumed both as
special cases.  We provided methods for obtaining various bounds from
different rules and also proved a generalization for implication
chains that determines the bounds for mixed rule-based inference
chains.  
An interesting extension of this would be to consider a system of nodes, each holding a probability that is updated in real-time as new information becomes available, and to then study how changes in the bounds propagate through the system. 
Furthermore, 
while the goal of this paper was not to speculate about
interpretations of the $\gamma$ parameter of the generalized rule, it
would be interesting to see whether there is a natural interpretation
of $\gamma$.  We could only briefly allude to %one way of 
thinking about it in terms of 
conditional probabilities, or in terms of
empirical support, i.e., the amount of data
available for deriving probabilistic rules, but other interpretations
are certainly possible and would be interesting to pursue.


\begin{acknowledgements} % will be removed in pdf for initial submission,
						 % (without ‘accepted’ option in \documentclass)
                         % so you can already fill it to test with the
                         % ‘accepted’ class optiona
%    Briefly acknowledge people and organizations here.
  %    \emph{All} acknowledgements go in this section.

This work was in part funded by ONR grant N00014-19-C-2044, AFOSR
grant FA9550-18-1-0465 and DARPA grant W911NF-20-2-0006.
  
\end{acknowledgements}

\bibliography{jahn-480.bib}

\end{document}














%\comment[MJ]{All of Section~\ref{sec:bounds_b} feels like an initial example, maybe should come earlier? 
%   The discussion references bounds on m,c,a,x so could not go before that?} 

\subsection{Bounds on $b$ (\modusponens)}  
\label{sec:bounds_on_b}
\paragraph{An example with bounds on $b$}

It will be helpful for understanding the motivation behind using
probability intervals for rule-based systems to use a running
example. Consider the sample space $\Omega = \set{1, 2, 3, 4}$ with
the three sets (``propositions'') $A = \set{1, 2}$, $B = \set{1, 3,
4}$, and $D = \set{2, 4}$ with their associated probabilities $1/2$,
$3/4$, and $1/2$ (assuming a uniform distribution). We can now define both
material and Bayesian conditionals:
\begin{alignat*}{3}
  &\Pr(A \rightarrow B) & &=\Pr(\{1, 3, 4\})            & &=3/4; \notag \\
  &\Pr(B \given A)      & &=\Pr(\{1\})/\Pr(\{(1, 2\})   & &=1/2; \notag \\
  &\Pr(B \rightarrow D) & &=\Pr(\{2, 4\})               & &=1/2; \notag \\
  &\Pr(D \given B)      & &=\Pr(\{4\})/\Pr(\{1, 3, 4\}) & &=1/3; \notag \\
  &\Pr(A \rightarrow D) & &=\Pr(\{2, 3, 4\})            & &=3/4; \notag \\
  &\Pr(D \given A)      & &=\Pr(\{2\})/\Pr(\{1, 2\})    & &=1/2.
\end{alignat*}

%\comment[MJ]{break into two subexamples, one with precise A, the other with bounded A.
%It is the uncertainty in $a$ that leads to different bounds wrt $m$ vs $c$.}

We can now compare the inferences using material and Bayesian \modusponens. 
Material \modusponens applied with $\Pr(A \impl B)=3/4$ and $\Pr(A) = 1/2$ yields $\Pr(B)  \in [1/4, 3/4]$,
and so does Bayesian \modusponens when applied with $\Pr(B\given A)=1/2$ and $\Pr(A)=1/2$.  
Similarly, applying material \modusponens with $\Pr(B \impl D)=1/2$ and $\Pr(B) = 3/4$ yields $\Pr(D) \in [1/4, 1/2]$,
and again the corresponding Bayesian \modusponens gives the same result. % 1/3*3/4,1/3*3/4 + 1 - 3/4

We now compare the two inferences when the antecedent is not precise. 
Suppose we did not know $\Pr(B)$, but only had the bounds computed above, $\Pr(B)\in[1/4,3/4]$.
Material \modusponens applied with $\Pr(B \impl D)=1/2$ and $\Pr(B) \in [1/4,3/4]$ 
% [1/4,3/4] -> max(1/4 + 1/2 - 1, 0), 1/2     ==> [0,1/2]
yields $\Pr(D)\in[0, 1/2]$, while Bayesian \modusponens applied with $\Pr(D \given B)=1/3$ and $\Pr(B)\in[1/4,3/4]$
% [1/4,3/4]  \given   1/4*1/3, 3/4 * 1/3 + 1 - 3/4   ==> [1/12,1/2]
yields $\Pr(D)\in[1/12,1/2]$.  
%
Now, for the sake of comparison, suppose we instead had $\Pr(B)\in[1/4,1/2]$, but the other probabilities
are the same as above. 
Then material \modusponens still gives $\Pr(D)\in[0,1/2]$, but
% 1/4 * 1/3, 1/2 * 1/3 + 1 - 1/2 = 1/6 + 3/6 = 4/6 => [1/12,2/3]
we get $\Pr(D)\in[1/12,2/3]$ from Bayes. 
%Thus the intersection would have yielded $\Pr(D)\in[1/12,1/2]$, a properly contained bound, 
%better than either one obtained via either material or Bayesian conditional inference given $\Pr(A)$. 

From these examples it seems 
that Bayes gives a better lower bound, that material gives a better upper bound,
that Bayesian \modusponens might give tighter bounds, 
     and that we should combine Bayes and material \modusponens to get optimal bounds.
The general analysis below shows that the first two are true, and the last two are misleading
or false.    

%\comment[MJ]{Note: $\Delta b(a,m)=6/12$, while $\Delta b(a,c)=9/12$. 
%    These are different because of clipping, i.e., 
%${b}^m \in [ 1/4 + 1/2 -1 = -1/4 \doteq 0, \ \ 1/2]$ and 
%${b}^c \in [ 1/12, \ \ 1-1/4(2/3)=10/12]$. 
%So without clipping, $\Delta b = 9/12$ for both.
%}

%The reason for the different bounds above corresponds to the simple observation that 
%you get better bounds if you compute with precise, rather than bounded, quantities.
%For example, if we know $m$ precisely, then material \modusponens gives
%$\L{b} = \L{x}(a,m) = m+\L{a}-1$. 
%This is algebraically equal to $\L{a}\L{c}$ by Lemma~\ref{lem:bounds},
%so if we compute using $m$, and not $c$ directly, we get $\L{x}=\L{a}\L{c} \le \L{a}c$.

We compute only the lower bounds in the comparisons below, but the upper bounds are similar.
If both $m$ and $c$ are precise, then $a$ is also precise. Both Material and Bayes give
$\L{x}(a,m) = a+m-1 = ac$. Now we compare the cases when only one of $m$ or $c$ is precise.
%
First, assume precise $m$, and computed $c$, i.e., 
$m =\L{m}=\U{m}$,  $\L{c} = (m+\L{a}-1)/\L{a}$. 
Then material \modusponens gives  $\L{x}(a,m) = \L{a}+m-1 = \L{a}\L{c}$ %\le \text{ Bayes}
and Bayesian \modusponens gives $\L{x}(a,c) = \L{a}\L{c} %\le \text{ precise Bayes}
\le \L{a}{c}$ ($=$ Bayes with precise $c$).
%
Next, assume precise $c$ and computed $m$, i.e., $\L{c} = \U{c} = c$,  $\L{m} = 1-\U{a}(1-c)$.
Then material \modusponens gives  $\L{x}(a,m)   = \L{a}+\L{m}-1 = \L{a}-\U{a}(1-c) 
                                  \le \L{a}+m-1$ ( = Material with precise $m$).
and Bayes gives $\L{x}(a,c)   = \L{a}{c}$.
%\begin{align*}
%&& m &=\L{m}=\U{m} 
%\\
%&& \L{c} &= (m+\L{a}-1)/\L{a} 
%\\
%&\text{Material w/ precise $m$:}   & \L{x}(a,m)    &= \L{a}+m-1 = \L{a}\L{c} %\le \text{ Bayes}
%\\
%&\text{Bayes w/ computed $c$:}     & \L{x}(a,c)    &= \L{a}\L{c} %\le \text{ precise Bayes}
%\\                                  &&              &\le \L{a}{c} \text{ = Bayes w/ precise $c$}
%\intertext{So Material gives the same value with either precise or computed data.
%    Next, assume precise $c$ and computed $m$.} 
%%Then  $\L{c}=\U{c} = c \ \ \Impl \ \ \L{m} = 1-\U{a}(1-c)$.}
%&& \L{c} &= \U{c} = c 
%\\
%&& \L{m} &= 1-\U{a}(1-c)
%\\
%&\text{Material w/ computed $m$:}  &  \L{x}(a,m)   &= \L{a}+\L{m}-1 = \L{a}-\U{a}(1-c) 
%\\                                  &&              &\le \L{a}+m-1 \text{ = Material w/ precise $m$}
%\\
%&\text{Bayes w/ precise $c$:}      &  \L{x}(a,c)   &= \L{a}{c} 
%\end{align*}

So Bayes (with precise $c$) gives better lower bounds. Similarly, material (with precise $m$) 
gives better upper bounds. 
%The issue is not approximating $x$ with approximate data,
%but maybe approximating $x$ with different functions.
%Maybe has to do with partials of $m$, $c$ wrt $a$ having different sign(?)
%
Note however, if $\bounds{a}$ is non-trivial, then at least one of $m$, $c$ must have non-trivial bounds 
(since if both $m$, $c$ are precise, then we know $a$).
Hence, it does not make sense to compute the lower bound with $\bounds{a},c$ (i.e., Bayes), 
and the upper bound with $\bounds{a},m$ (i.e., Material), 
and then to \textit{combine} them to get better bounds on $d$. 
The point is that if we have both $m$, $c$ precise, then $a$ is also precise, and both computations
for $\L{x}$ will give the same result (and similarly for $\U{x}$).  
Moreover, if $\L{a}<a$ (where $a$ is computed from $m$, $c$, and $\L{a}$ is the original, 
not-updated lower bound) then the computation for $\L{x}$ will improve, since $\L{a}c < ac$,
and this will yield a better lower bound than the ``combined'' approach above.
%
Furthermore, the width of the intervals from both Bayes and material are the same ($=1-\L{a}$).
The reason that one might give a narrower interval is because 
the computed bounds %values of $\L{a}$, $\U{a}$ 
might be outside $[0,1]$, so when we apply Definition~\ref{def:LU} we get a shorter interval.

Now we want to compute bounds on $b$ when we have a generalized implication $A \gimpl B$.
The equations in Lemma~\ref{lem:fn(mcax)} are true for any $B$ with $\Pr(B \given A)=c$,
but $b$ is itself not functionally related to any of $m,c,a$.
Consequently, we can only get bounds on $b$, even if we know all of $m,c,a,x$ precisely.
In this case, we use
$\Pr(A \intersect B) \le \Pr(B) \le \Pr(\comp{A} \union B)$, to get the well-known (clearly tight) bounds
$x \le b \le m$, i.e.,
\begin{align}
\label{eqn:bLbU}
\L{b} &= {x} \ , \quad \U{b} = {m} .
\end{align}
Using Lemma~\ref{lem:fn(g):x},~\ref{lem:fn(g):m}, we can express these bounds in terms of $g$:
\begin{align}
\L{b}   &=(1-\gamma){a}{g} + \gamma({a}+{g}-1)              
\label{eqn:bL(a,g)}
\\
\U{b}   &= (1-\gamma){a}{g} + \gamma({a}+{g}-1) + (1-{a}) . 
\label{eqn:bU(a,g)}
\end{align}
The reader might note that these expressions are just a special case of
Lemma~\ref{lem:fn(g):ghat}.
Of course, these formulas apply when we have precise probabilities for $a,g$.
    If we only have bounds for $a$ and $g$, we compute 
    $\dfrac{\partial}{\partial a}\L{b}(a,g)>0$,\
$\dfrac{\partial}{\partial g}\L{b}(a,g)>0$, and 
    $\dfrac{\partial}{\partial a}\U{b}(a,g)<0$,\ 
    $\dfrac{\partial}{\partial g}\U{b}(a,g)>0$.
Using Lemma~\ref{lem:bounds} we can compute $\L{(\L{b})}=\L{x}$, $\U{(\L{b})}=\U{x}$, $\L{(\U{b})}=\L{m}$, $\U{(\U{b})}=\U{m}$.
We will abuse notation and write $\L{b}=\L{(\L{b})}$ and $\U{b}=\U{(\U{b})}$, and discard the other two:
\begin{align}
\L{b}   &= \L{b}(\L{a},\L{g})
        = (1-\gamma)\L{a}\L{g} + \gamma(\L{a}+\L{g}-1)             
\label{eqn:bL(aL,gL)}
\\ 
\U{b}   &= \U{b}(\L{a},\U{g})
        = (1-\gamma)\L{a}\U{g} + \gamma(\L{a}+\U{g}-1) + (1-\L{a}) . 
\label{eqn:bU(aL,gU)}
\end{align}
Whether $\L{b}$ refers to the function in (\ref{eqn:bL(a,g)}) or the value in (\ref{eqn:bL(aL,gL)})
will be clear from context.  Furthermore, 
$\Delta{b} = (1-\L{a}) + [(1-\gamma)\L{a}+\gamma\!\cdot\!1]\Delta g$, and 
if $g$ is precise we just have $\Delta{b}=1-\L{a}$, as noted above for the case
of precise $m$, $c$. 

%\comment[fixed]{Comment:  (\ref{eqn:bL(a,g)}, \ref{eqn:bU(a,g)}) are really just new equations, 
%so we should be computing $\L{(\L{b})}, \U{(\U{b})}$ in (\ref{eqn:bL(aL,gL)},\ref{eqn:bU(aL,gU)}).
%The point is that $\L{b}$ has bounds $\bounds{(\L{b})}=\bounds{x}$. 
%We will just write $\L{b}$ in place of $\L{(\L{b})}$ for the sake of simplicity.
%See Section~\ref{sec:converses} - Converses for more explanation.}

%\markblue
%To decrease $\Delta b$ we need change $\gamma$ in order to decrease the product
%    $[(1-\gamma)\L{a}+\gamma\!\cdot\!1]\Delta g$.  
%    \comment[MJ]{See\ section~\ref{sec:observation_on_partial_g} ??}
%    If $\L{a}=\U{a}=a$, then \ref{lem:bounds:g(a,c)} gives
%    $[(1-\gamma)\L{a}+\gamma\!\cdot\!1] \Delta g = a\Delta c$. 
%    Similarly, \ref{lem:bounds:g(a,m)} gives
%    $[(1-\gamma)\L{a}+\gamma\!\cdot\!1] \Delta g = \Delta m$.
%    So in either of these cases we get %(\ref{def:ADg}) is
%    \begin{align*}
%    \frac{\partial}{\partial \gamma}\Big([(1-\gamma)\L{a}+\gamma\!\cdot\!1] \Delta g\Big) &= 0
%    \end{align*}
%    and changing $\gamma$ won't help.
%    \comment[MJ]{Numerical computations indicate $\ne0$ when $\Delta a\ne0$, but
%        that monotonicity implies $\gamma=1$  gives smallest $\Delta b$ when based on $\gamma=0$
%            and $c$ is precise. ??? }
%
%\marknormal








