08-conditionals.tex

\chapter{Conditionals}\label{ch:conditionals}

\section{Material conditionals}\label{sec:material}

We are often interested not just in whether something is in fact the case, but
also in whether it is (or would be) the case \emph{if} something else is (or
would be) the case. We might, for example, wonder in what will happen to the
climate if we don't reduce greenhouse gases, or whether World War 2 could have
been avoided if certain steps had been taken in the 1930s.

A sentence stating that something is (or would be) the case if something else is
(or would be) the case is called a \textbf{conditional}. What exactly, do these
statements mean? What is their logic? Philosophers have puzzled over these
questions for more than 2000 years, with no agreement in sight.

One attractively simple view is that a conditional `if $A$ then $B$' is true iff
the antecedent $A$ is false or the consequent $B$ is true. This would make `if
$A$ then $B$' equivalent to `not $A$ or $B$'. Conditionals with these
truth-conditions are called \textbf{material conditionals}.

The conditionals $A \to B$ of classical logic are material. $A \to B$ is
equivalent to $\neg A \lor B$. The attractively simple view that English
conditionals are material conditionals would mean that we can faithfully
translate English conditionals into $\L_{M}$-sentences of the form $A \to B$. Is
this correct?

There are some arguments for a positive answer. Suppose I make the following
promise.
\begin{itemize}[leftmargin=10mm]
  \item[(1)] If I don't have to work tomorrow then I will help you move.
\end{itemize}
I have made a false promise if the next day I don't have to work and yet I don't
help you move. Under all other conditions, you could not fault me for breaking my
promise. So it seems that (1) is false iff I don't have to work and I don't help
you move. Generalizing, this suggests that `if $A$ then $B$' is true iff $A$ is
false or $B$ is true.

Another argument for analysing English conditionals as material conditionals
starts with the intuitively plausible assumption that `$A$ or $B$' entails the
corresponding conditional `if not-$A$ then $B$'. (This is sometimes called the
\emph{or-to-if} inference.) Suppose I tell you that Nadia is either in Rome or
in Paris. Trusting me, you can infer that if she's not in Rome then she's in
Paris. Now we can reason as follows.

To begin, suppose that $A$ and `if $A$ then $B$' are both true. Plausibly, we
can infer that $B$ is true as well: \emph{modus ponens} is valid for English
conditionals. \label{p:mp} This means that if $A$ is true and $B$ is false, then
`if $A$ then $B$' is false. Now suppose, alternatively, that $A$ is false or $B$
is true. Then `not-$A$ or $B$' is true. By or-to-if, we can infer that `if $A$
then $B$' is true as well. Thus `if $A$ then $B$' is true iff $A$ is false or
$B$ is true.

Despite these arguments, most philosophers and linguists don't think that
English conditionals are material conditionals. Consider these facts about
logical consequence (in classical propositional logic).
%
\label{paradoxes-mat-imp}
\begin{principles}
\pri{M1}{B \models A \to B}\\
\pri{M2}{\neg A \models A \to B}\\
\pri{M3}{\neg (A \to B) \models A}\\
%\item (A \to B) \lor (B \to A)}\\
%\pri (A \to B) \lor (B \to C)}\\
\pri{M4}{A \to B \models \neg B\to \neg A}\\
\pri{M5}{A \to B \models (A\land C) \to  B}
\end{principles}

If English conditionals were material conditionals then the following
inferences, corresponding to (M1)--(M5), would be valid.

\begin{enumerate}[leftmargin=12mm]
  \itemsep1mm
  \item[(E1)] There won't be a nuclear war. Therefore: If Russia attacks the US
        with nuclear weapons then there won't be a nuclear war.
  \item[(E2)] There won't be a nuclear war. Therefore: If there will be a
        nuclear war then nobody will die.
  \item[(E3)] It is not the case that if it will rain tomorrow then the Moon
        will fall onto the Earth. Therefore: It will rain tomorrow.
  \item[(E4)] If our opponents are cheating, we will never find out. Therefore:
        If we will find out that our opponents are cheating, then they aren't
        cheating.
  \item[(E5)] If you add sugar to your coffee, it will taste good. Therefore: If
        you add sugar and vinegar to your coffee, it will taste good.
\end{enumerate}

These inferences do not sound good. If we wanted to defend the view that English
conditionals are material conditionals we would have to explain why they sound
bad even though they are valid. We will not explore this option any further.

\begin{exercise}
  Can you find a different analysis of English conditionals that, like the
  material analysis, would make conditionals truth-functional, but that would
  render all of (E1)--(E5) invalid?
\end{exercise}
\begin{solution}
  (E1)--(E5) are invalid assuming that `if $A$ then $B$' is true iff both $A$
  and $B$ are true. There are, of course, strong reasons against the analysis of
  English conditionals as conjunctions.
\end{solution}

Even those who defend the material analysis of English conditionals admit that
it does not work for all English conditionals. Consider (2).
\begin{enumerate}[leftmargin=10mm]
\item[(2)] If water is heated to $100^\circ$ C, it evaporates.
\end{enumerate}
This shouldn't be translated as $p\to q$. Intuitively, (2) states that \emph{in
  all (normal) cases} where water is heated to $100^\circ$ C, it evaporates. It
is a quantified, or modal claim.

Another important class of conditionals that can't be analysed as material
conditionals are so-called \textbf{subjunctive conditionals}. Compare the
following two statements.

\begin{enumerate}[leftmargin=10mm]
  \itemsep-1mm
\item[(3)] If Shakespeare didn't write \emph{Hamlet}, then someone else did.
\item[(4)] If Shakespeare hadn't written \emph{Hamlet}, then someone else
  would have.
\end{enumerate}
%
(3) seems true. Someone has written \emph{Hamlet}; if it wasn't Shakespeare then
it must have been someone else. But (4) is almost certainly false. After all, it
is very likely that Shakespeare did write \emph{Hamlet}. And it is highly
unlikely that if he hadn't written \emph{Hamlet} -- if he got distracted by
other projects, say -- then someone else would have stepped in to write the
exact same piece.

Sentences like (3) are called \textbf{indicative conditionals}. Intuitively, an
indicative conditional states that something is \emph{in fact} the case on the
assumption that something else is the case. A subjunctive conditional like (4)
states that something \emph{would be} the case if something else \emph{were} the
case. Typically we know that the ``something else'' is not in fact the case. We
know, for example, that Shakespeare wrote \emph{Hamlet} and therefore that the
antecedent of (4) is false. For this reason, subjunctive conditionals are also
called \emph{counterfactual conditionals} or simply \emph{counterfactuals}.

It should be clear that subjunctive conditionals are not material conditionals.
I said that (4) is almost certainly false. But it almost certainly has a false
antecedent. So the corresponding material conditional is almost certainly true.

\section{Strict conditionals}\label{sec:strict-implication}

% On the history see http://hume.ucdavis.edu/mattey/phi134/strict.htm

One apparent difference between material conditionals $A \to B$ and conditionals
in natural language is that $A\to B$ requires no connection between the
antecedent $A$ and the consequent $B$. Consider (1).
\begin{enumerate}[leftmargin=10mm]
  \item[(1)] If we leave after 5, we will miss the train.
\end{enumerate}
Intuitively, someone who utters (1) wants to convey that missing the train
is a \emph{necessary consequence} of leaving after 5 -- that it is \emph{impossible} to
leave after 5 and still make it to the train, given certain facts about the
distance to the station, the time it takes to get there, etc. This suggests that
(1) should be formalized not as $p \to q$ but as $\Box(p \to q)$ or,
equivalently, $\neg \Diamond(p \land \neg q)$.

Sentences that are equivalent to $\Box(A \to B)$ are called \textbf{strict
  conditionals}. The label goes back to C.I.\ Lewis (1918), who also introduced
the abbreviation $A \strictif B$ for $\Box(A \to B)$.

Lewis was not interested in `if \ldots then \ldots' sentences. He introduced
$A \strictif B$ to formalize `$A$ implies $B$' or `$A$ entails $B$'. His
intended use of $\strictif$ roughly matches our use of the double-barred
turnstile `$\models$'. But there are important differences. The turnstile is an
operator in our \emph{meta-language}; Lewis's $\strictif$ is an
\emph{object-language} operator that, like $\land$ or $\to$, can be placed
between any two sentences in a formal language to generate another sentence in
the language. $p \strictif (q \strictif p)$ is well-formed, whereas
$p \models (q\models p)$ is gibberish. Moreover, while $p \models q$ is simply
false -- because there are models in which $p$ is true and $q$ false -- Lewis's
$p \strictif q$ is true on some interpretation of the sentence letters and false
on others. If $p$ means that it raining heavily and $q$ that it is raining, then
$p \strictif q$ is true because the hypothesis that it is raining heavily
implies that it is raining.

Let's set aside Lewis's project of formalizing the concept of implication. Our
goal is to find an object-language construction that functions like `if \ldots
then \ldots' in English. To see whether `$\ldots \strictif\ldots$' can do the job,
let's have a closer look at the logic of strict conditionals.

Since $A \strictif B$ is equivalent to $\Box(A \to B)$, standard Kripke
semantics for the box also provides a semantics for strict conditionals. In
Kripke semantics, $\Box(A \to B)$ is true at a world $w$ iff $A \to B$ is true
at all worlds $v$ accessible from $w$. And $A \to B$ is true at $v$ iff 
$A$ is false at $v$ or $B$ is true at $v$. We therefore have the following
truth-conditions for strict conditionals.

\begin{definition}{Kripke semantics for $\strictif$}{semstrictif}
  If $M = \t{W,R,V}$ is a Kripke model, then\\[1mm]
  $M,w \models A \strictif B$ iff for all $v$ such that $wRv$, either
  $M,v \not\models A$ or $M,v \models B$.
\end{definition}

\begin{exercise}
  $A \strictif B$ is equivalent to $\Box(A \to B)$. Can you fill the blank in:
  `$\Box A$ is equivalent to ---', using no modal operator other than
  $\strictif$?
\end{exercise}
\begin{solution}
  For example: $\neg A \strictif A$ or $(A\lor \neg A) \strictif A$.
\end{solution}

As always, the logic of strict conditionals depends on what constraints we
put on the accessibility relation. Without any constraints, $\strictif$ does
not validate \emph{modus ponens}, in the sense that $A \strictif B$ and $A$
together do not entail $B$. We can see this by translating
$A \strictif B$ back into $\Box(A \to B)$ and setting up a tree. Recall that to
test whether some premises entail a conclusion, we start the tree with the
premises and the negated conclusion.
%
\begin{center}
  \tree[3]{%
    & \nnode{18}{1.}{$\Box (A \to B)$}{w}{(Ass.)} &\\
    & \nnode{18}{2.}{$A$}{w}{(Ass.)} &\\
    & \nnode{18}{3.}{$\neg B$}{w}{(Ass.)} &
  }
\end{center}
%
With the K-rules, where we don't make any assumptions about the accessibility
relation, node 1 can't be expanded, so there is nothing more we can do.

\begin{exercise}
  Give a countermodel in which $p \strictif q$ and $p$ are true at some world
  while $q$ is false.
\end{exercise}
\begin{solution}
  $W = \{ w \}$, $R = \emptyset$, $V(p) = \{ w \} $, $V(q)=\emptyset$.
\end{solution}

If we assume that the accessibility relation is reflexive, the tree closes:
\begin{center}
  \tree[3]{%
    & \nnode{18}{4.}{$wRw$}{}{(Ref.)} &\\
    & \bnode{18}{5.}{$A \to B$}{w}{(1,4)} &\\
    &&\\
    \nnodeclosed{10}{6.}{$\neg A$}{w}{(5)} && \nnodeclosed{10}{7.}{$B$}{w}{(5)}
  }
\end{center}

It is not hard to show that \emph{modus ponens} for $\strictif$ is valid on all
and only the reflexive frames. Reflexivity is precisely what we need to render
\emph{modus ponens} valid. Since \emph{modus ponens} looks plausible for English
conditionals (as I've argued on p.~\pageref{p:mp}), we'll probably want the
relevant Kripke models to be reflexive.

\begin{exercise}\label{ex:sda-import}
  Using the tree method, and translating $A \strictif B$ into $\Box(A \to B)$, confirm that following claims hold, for all $A,B,C$.
  \begin{exlist}
  \item $\models_K A \strictif A$
  \item $A \strictif B \models_K \neg B \strictif \neg A$
  \item $A \strictif B \models_K (A \land C) \strictif B$
  \item $A\strictif B, B \strictif C \models_{K} A \strictif C$
  \item $(A \lor B) \strictif C \models_K (A \strictif C) \land (B \strictif C)$
  \item $A \strictif (B \strictif C) \models_T (A \land B) \strictif C$
%  \item $A \strictif B, \neg B \models_T \neg A$ 
  \item $A\strictif B \models_{S4} C \strictif (A \strictif B)$
    % inspired by Hacking's formula for 4, which is the necessitated version of the corresponding conditional
  \item $((A\strictif B) \strictif C) \strictif (A\strictif B) \models_{S5} A\strictif B$
  \end{exlist}
  \medskip

  Which of these schemas do you think should be valid if we
  assume that $A \strictif B$ translates indicative conditionals `if $A$ then $B$'?
\end{exercise}
\begin{solution}
  Use \href{https://www.umsu.de/trees/}{umsu.de/trees/}.
\end{solution}

% Note: if epistemic, then maybe S4.2; if subjunctive, then maybe 5??

We could now look at other conditions on the accessibility relation and decide
whether they should be imposed, based on what they would imply for the logic of
conditionals. But let's take a shortcut.

I have suggested that sentence (1) might be understood as saying that it is
\emph{impossible} to leave after 5 and still make it to the train. Impossible in
what sense? There are many possible worlds at which we leave after 5 and still
make it to the train. There are, for example, worlds at which the train departs
two hours later, worlds at which we live right next to the station, and so on.
When I say that it is impossible to leave after 5 and still make it to the
train, I arguably mean that it is impossible \emph{given what we know about the
  departure time, our location, etc.}

Generalizing, a tempting proposal is that the accessibility relation that is
relevant for indicative conditionals like (1) is the epistemic accessibility
relation that we studied in chapter \ref{ch:epistemic}, where a world $v$ is
accessible from $w$ iff it is compatible with what is known at $w$. On that
hypothesis, the logic of indicative conditionals is determined by the logic
of epistemic necessity. We don't need to figure out the relevant accessibility
relation from scratch.

Since knowledge varies from agent to agent, the present idea implies that the
truth-value of indicative conditionals should be agent-relative. This seems to
be confirmed by the following puzzle, due to Allan Gibbard.
\begin{quote}
  Sly Pete and Mr.\ Stone are playing poker on a Mississippi riverboat. It is
  now up to Pete to call or fold. My henchman Zack sees Stone’s hand, which is
  quite good, and signals its content to Pete. My henchman Jack sees both hands,
  and sees that Pete's hand is rather low, so that Stone's is the winning hand.
  At this point the room is cleared. A few minutes later, Zack slips me a note
  which says `if Pete called, he won', and Jack slips me a note which says `if
  Pete called, he lost'. % \cite[231]{gibbard1981xxx}
\end{quote}
The puzzle is that Zack's note and Jack's note are intuitively contradictory,
yet they both seem to be true.

We can resolve the puzzle if we understand the conditionals as strict
conditionals with an agent-relative epistemic accessibility relation. Take Zack.
Zack knows that Pete knows Stone's hand. He also knows that Pete would not call
unless he has the better hand. So among the worlds compatible with Zack's
knowledge, all worlds at which Pete calls are worlds at which Pete wins. If $p$
translates `Pete called' and $q$ `Pete won', then $p \strictif q$ is true
relative to Zack's information state. Relative to Jack's information state,
however, the same sentence is false. Jack knows that Stone's hand is better than
Pete's, but he doesn't know that Pete knows Stone's hand. Among the worlds
compatible with Jack's knowledge, all worlds at which Pete calls are therefore
worlds at which Pete loses. Relative to Jack's information state,
$p \strictif \neg q$ is true.

Another advantage of the ``epistemically strict'' interpretation is that it
might explain why indicative conditionals with antecedents that are known to be
false seem defective. For example, imagine a scenario in which Jones has gone to
work. In that scenario, is (2) true or false?
\begin{enumerate}[leftmargin=10mm]
  \item[(2)] If Jones has not gone to work then he is helping his neighbours.
\end{enumerate}
The question is hard to answer -- and not because we lack information about the
scenario. Once we are told that Jones has gone to work, it is unclear how we are
meant to assess whether Jones is helping his neighbours \emph{if} he has not
gone to work. On the epistemically strict interpretation, (2) says that Jones is
helping his neighbours at all epistemically accessible worlds at which Jones
hasn't gone to work. Since we know that Jones has gone to work, there are no
epistemically accessible worlds at which he hasn't gone to work. And if there
are no $A$-worlds then we naturally balk at the question whether all $A$-worlds
are $B$-worlds. (In logic, we resolve to treat `all $A$s are $B$' as true if
there are no $A$s. Accordingly, (2) comes out true on the epistemically strict
analysis. But we can still explain why it seems defective.)

We have found a promising alternative to the hypothesis that indicative
conditionals are material conditionals. According to the present alternative,
they are epistemically strict conditionals -- strict conditionals with an
epistemic accessibility relation.

What about subjunctive conditionals? Return to the two Shakespeare conditionals
from the previous section. When we evaluate the indicative sentence -- `If
Shakespeare didn't write \emph{Hamlet}, then someone else did' -- we hold fixed
our knowledge that \emph{Hamlet} exists; worlds where the play was never written
are inaccessible. That's why the conditional is true. At all accessible worlds
at which Shakespeare didn't write \emph{Hamlet}, someone else wrote the play.
When we evaluate the subjunctive conditional -- `If Shakespeare hadn't written
\emph{Hamlet}, then someone else would have' -- we do consider worlds at which
\emph{Hamlet} was never written, even though we know that the actual world is
not of that kind. If subjunctive conditionals are strict conditionals, then
their accessibility relation does not track our knowledge or information.
Unfortunately, as we are going to see in the next section, it is hard to say
what else it could track.

This is one problem for the strict analysis of natural-language conditionals.
Another problem lies in the logic of strict conditionals. Remember (E1)--(E5)
from page \pageref{paradoxes-mat-imp}. If English conditionals are strict
conditionals, then (E1)--(E3) are invalid. For example, while $q$ entails
$p \to q$, it does not entail $p \strictif q$. But the strict analogs of
\pr{M4} and \pr{M5} still hold, no matter what we say about accessibility (see
exercise \ref{ex:sda-import}):
\begin{gather*}
  A \strictif B \models \neg B\strictif \neg A;\\
  A \strictif B \models (A\land C) \strictif  B.
\end{gather*}
So we still predict that the inferences (E4) and (E5) are valid.

\begin{enumerate}[leftmargin=12mm]
  \itemsep1mm
  \item[(E4)] If our opponents are cheating, we will never find out. Therefore:
        If we will find out that our opponents are cheating, then they aren't
        cheating.
  \item[(E5)] If you add sugar to your coffee, it will taste good. Therefore: If
        you add sugar and vinegar to your coffee, it will taste good.
\end{enumerate}

\begin{exercise}
  The badness of (E4) and (E5) suggests that indicative conditionals can't be
  analysed as strict conditionals. Can you give a similar argument suggesting
  that \emph{subjunctive} conditionals can't be analysed as strict conditionals?
\end{exercise}
\begin{solution}
  (E1)--(E5) all work equally well in the subjunctive mood. For \pr{E4} and
  \pr{E5}:
  \begin{itemize}
  \item If our opponents had been cheating, we would never have found out.
    Therefore: If we had found out that our opponents are cheating, then
    they wouldn't have been cheating.
  \item If you had added sugar to your coffee, it would have tasted
    good. Therefore: If you had added sugar and vinegar to your coffee, it would
    have tasted good.
  \end{itemize}
  Both of these inferences are valid if subjunctive conditionals are strict
  conditionals. But they don't sound good.
\end{solution}

% Another problem: subjunctive mights. Might if A then B doesn't seem to mean
% $\Diamond\Box(A \to B)$, nor $\Box(A \to \Diamond B)$.

% \begin{exercise}
%   Some have argued that problems similar to those raised by
%   \pr{M1}--\pr{M3} remain for strict implication, because the
%   following turn out to be true:
%   \begin{enumerate*}
%   \item $\Box B \models A \strictif B$
%   \item $\Box \neg A \models A \strictif B$
%   \item $\neg (A \strictif B) \models \Diamond A$
%   \end{enumerate*}
  
%   Note analogy to wide-scope analysis of conditional obligation.
% \end{exercise}

% Arguably, natural language conditionals lie in between strict and material. For
% if $A \to B$ is false, then $A$ is true and $B$ false, and then `if $A$ then
% $B$' is surely false. And if $A\to B$ is necessary, so at all worlds where $A$
% holds $B$ holds, then plausibly if $A$ holds (or were to hold) at the actual
% world, then so does (would) $B$.

% Every SC model (as defined here) is a VC model (as defined in next sec). So
% everything that's true in all VC models is true in all SC models; i.e. the
% logic SC is strictly stronger than VC.

\begin{exercise}
  A plausible norm of pragmatics is that a sentence should only be asserted if
  it is known to be true. Let's call a sentence \emph{assertable} if it is known
  to be true. Show that if the logic of knowledge is at least S4, then an
  epistemically strict conditional $A \strictif B$ is assertable iff the
  corresponding material conditional $A \to B$ is assertable.
\end{exercise}
\begin{solution}
  Suppose $A \to B$ is assertable. Then $A\to B$ is known. So $\Kn (A \to
  B)$. In S4, it follows that $\Kn\Kn(A \to B)$. So the epistemically strict
  conditional $\Kn(A \to B)$ is assertable. Conversely, if $\Kn(A \to B)$ is
  assertable, then it is known; so $\Kn\Kn(A \to B)$. In S4, it follows that
  $\Kn(A \to B)$. So $A \to B$ is assertable.
\end{solution}

\begin{exercise}
  Explain why the `or-to-if' inference from `$p$ or $q$' to `if not $p$ then
  $q$' is invalid on the assumption that the conditional is epistemically
  strict. How could a friend of this assumption explain why the inference
  nonetheless looks reasonable, at least in normal situations? (Hint: Remember
  the previous exercise.)
\end{exercise}
\begin{solution}
  The `or-to-if' inference is not valid on the assumption that the conditional
  is epistemically strict. For example, if $p$ and $q$ are both true at the actual world and both false at some epistemically accessible world, then `$p$ or $q$' is true but `if $p$ then $q$' is false (on the strict analysis).

  The inference might nonetheless look reasonable because it would normally be
  inappropriate to assert a disjunction `$p$ or $q$' unless the disjunction is
  known -- unless it is true at all epistemically accessible worlds. And if
  $p \lor q$ is true at all epistemically accessible worlds then $\neg p \to q$
  is also true at all epistemically accessible worlds, and so $\Box(\neg p \to q)$ is
  true. Thus the conclusion of or-to-if is true in any situation in which the
  premise is \emph{assertable}. If the logic of knowledge validates the
  \pr{4}-schema, we can go further and say that the conclusion is assertable in
  any situation in which the premise is assertable.
\end{solution}

  
\section{Variably strict conditionals}

Let's have a closer look at subjunctive conditionals. As I am writing these
notes, I am sitting in Coombs Building, room 2228, with my desk facing
the wall to Al H\'ajek's office in room 2229. In light of these facts, (1) seems
true.
\begin{enumerate}[leftmargin=10mm]
\item[(1)] If I were to drill a hole through the wall behind my desk,
  the hole would come out in Al's office.
\end{enumerate}
% This example is meant to get around some of Al's worries.
There is no logical connection between the antecedent of (1) and the consequent.
There are many possible worlds at which I drill a hole through the wall behind
my desk and don't reach Al's office -- for example, worlds at which my desk
faces the opposite wall, worlds at which Al's office is in a different room, and
so on. If (1) is a strict conditional then all such worlds must be inaccessible.

Now consider (2).
\begin{enumerate}[leftmargin=10mm]
\item[(2)] If the office spaces had been randomly reassigned yesterday then Al's
  office would (still) be next to mine.
\end{enumerate}
(2) seems false, or at least very unlikely. But if (2) is a strict conditional,
and worlds at which Al is not in room 2229 or I am not in 2228 are inaccessible
-- as they seem to be for (1) -- then (2) should be true. Among worlds at which
I am in 2228 and Al is in 2229, all worlds at which the office spaces have been
randomly reassigned yesterday are worlds at which Al's office is next to mine.
When we evaluate (2), it looks like we no longer hold fixed who is in which
office. Worlds that were inaccessible for (1) are accessible for (2).

So the accessibility relation, at least for subjunctive conditionals, appears to
vary from conditional to conditional. As David Lewis put it, subjunctive
conditionals seem to be not strict, but ``variably strict''.

Let's try to get a better grip on how this might work. (What follows is a
slightly simplified version of an analysis developed by Robert Stalnaker and
David Lewis in the 1960s.)

Intuitively, when we ask what would have been the case if a certain event had
occurred, we are looking at worlds that are much like the actual world up to the
time of the event. Then these worlds deviate in some minimal way to allow the
event to take place. Afterwards the worlds unfold in accordance with the general
laws of the actual world.

For example, if we wonder what would have happened if Shakespeare hadn't written
\emph{Hamlet}, we are interested in worlds that are like the actual world until
1599, at which point some mundane circumstances prevent Shakespeare from writing
\emph{Hamlet}. We are not interested in worlds at which Shakespeare was never
born, or in which the laws of nature are radically different from the laws at
our world. One might reasonably judge that Shakespeare would have been a famous
author even if he hadn't written \emph{Hamlet}, although we would hardly be
famous in worlds in which he was never born.

Likewise for (1). Here we are considering worlds that are much like the actual
world up to now, at which point I decide to drill a hole and find a suitable
drill. These changes do not require my office to be in a different room. Worlds
where I'm not in room 2228 can be ignored. Figuratively speaking, such worlds
are ``too remote'': they differ from the actual world in ways that are not
required to make the antecedent true.

This suggests that a subjunctive conditional is true iff the consequent is true
at the ``closest'' worlds at which the antecedent is true -- where ``closeness''
is a matter of similarity in certain respects. The closest worlds (to the actual
world) at which Shakespeare didn't write \emph{Hamlet} are worlds that almost
perfectly match the actual world until 1599, then deviate a little so that
Shakespeare didn't write Hamlet, and afterwards still resemble the actual world
with respect to the general laws of nature. We will not try to spell out in full
generality what the relevant closeness measure should look like.

Let `$v \prec_w u$' mean that $v$ is closer to $w$ than $u$, in the sense that
$v$ differs less than $u$ from $w$ in whatever respects are relevant to the
interpretation of subjunctive conditionals.

We make the following structural assumptions about the world-relative ordering
$\prec$.

\begin{enumerate}[leftmargin=8mm]
  \itemsep-1mm
  \item If $v \prec_w u$ then $u \nprec_w v$. (Asymmetry)
  \item If $v \prec_w u$, then for all $t$ either $v \prec_w t$ or
        $t \prec_w u$. (Quasi-connectedness)
% \item For all $w$ and $v$, $v \nprec_w w$. (\textbf{Weak centring})
  \item For any non-empty set of worlds $X$ and world $w$ there is a $v$ in $X$
        such that there is no $u$ in $X$ with $u \prec_w v$.
  % There are different formulations of the Limit Assumption. This is what
  % Kaufmann 2017 calls PLA.
\end{enumerate}

Asymmetric and quasi-connected relations are known as \textbf{weak orders}.
Asymmetry is self-explanatory. Quasi-connectedness is more often called
\emph{negative transitivity}, because it is equivalent to the assumption that if
$t \not< s$ and $s\not<r$ then $t\not<r$. It ensures that the ``equidistance''
relation that holds between $v$ and $u$ if neither $v \prec_w u$ nor
$u \prec_w v$ is an equivalence relation. With these two assumptions, we can
picture each world $w$ as associated with nested spheres of worlds;
$v \prec_w u$ means that $v$ is in a more narrow $w$-sphere than $u$.

Assumption 3 is known as the \textbf{Limit Assumption}. It ensures that for any
consistent proposition $A$ and world $w$, there is a set of closest $A$-worlds.
Without the Limit Assumption, there could be an infinite chain of ever closer
$A$-worlds, with no world being maximally close.

% Maybe express similarity models in terms of spheres, like neighbourhood models,
% i.e. N(w) = { { w }, { w, v }, ... }? This makes the frames easier to draw and
% to write. Students don't know how to write that two worlds are tied in terms
% of <_{w}.

\begin{exercise}
  Show that asymmetry and quasi-connectedness imply transitivity.
\end{exercise}
\begin{solution}
  Assume that $R$ is asymmetric and quasi-connected. We want to show that
  $R$ is transitive. So assume we have $xRy$ and $yRz$. By quasi-connectedness,
  $yRz$ implies that either $yRx$ or $xRz$. By asymmetry, we can't have $yRx$,
  since we have $xRy$. So $xRz$.
\end{solution}

\begin{exercise}
  Define $\preceq_{w}$ so that $v \preceq_w u$ iff $u \nprec_w v$ (that is, iff
  it is not the case that $u \prec_{w} v$). Informally, $v \preceq_w u$ means
  that $v$ is at least as similar to $w$ in the relevant respects as $u$. Many
  authors use $\preceq$ rather than $\prec$ as their basic notion. Can you
  express the above three conditions on $\prec$ in terms of $\preceq$?
\end{exercise}
\begin{solution}
  We have the following equivalences (using `$\Leftrightarrow$' to mean that the
  expressions on either side are equivalent):
  \[
    u \npreceq_w v \Leftrightarrow \neg (u \preceq_w v) \Leftrightarrow \neg(v \nprec_w u) \Leftrightarrow v \prec_w u.
  \]
  So you can simply replace every instance of $\omega \prec_{w} \nu$ in the
  conditions by $\nu \npreceq_{w} \omega$, and every instance of
  $\omega \nprec_{w} \nu$ by $\nu \preceq_{w} \omega$.

  Asymmetry thereby turns into: if $u \npreceq_{w} v$ then $v \preceq_{w} u$.
  Equivalently: either $u \preceq_{w} v$ or $v \preceq_{w} u$. This
  property of relations is called \textbf{completeness}. Notice that it
  entails reflexivity.

  Quasi-connectedness turns into: if $u \npreceq_{w} v$ then for all $t$, either
  $t \npreceq_{w} v$ or $u \npreceq_{w} t$. This is equivalent to transitivity
  for $\preceq$.

  % Weak centring turns into: for all $w$ and $v$, $w \preceq_{w} v$.

  The Limit Assumption turns into: for any non-empty set of worlds $X$ and world
  $w$ there is a $v\in X$ such that there is no $u \in X$ with
  $v \npreceq_{w} u$. Equivalently, for any non-empty set of worlds $X$
  and world $w$ there is a $v\in X$ such that $v \preceq_{w} u$ for all $u\in X$.
\end{solution}

We are going introduce a variably strict operator $\boxright$ so that
$A\boxright B$ is true at a world $w$ iff $B$ is true at the closest worlds to
$w$ at which $A$ is true. Models for a language with the $\boxright$ operator
must contain closeness orderings $\prec$ on the set of worlds.

\begin{definition}{}{similaritymodel}
  A \textbf{similarity model} consists of
  \vspace{-3mm}
  \begin{itemize*}
  \item a non-empty set $W$,
  \item for each $w$ in $W$ a weak order $\prec_w$ that satisfies the Limit
  Assumption, and
  \item a function $V$ that assigns to each sentence letter a subset of $W$.
  \end{itemize*}
  % Notice that this is a special kind of neighbourhood model: the propositions
  % necessary at w are the nested spheres around w.
\end{definition}

To formally state the semantics of $\boxright$, we can re-use a concept from
section \ref{sec:oblig-circ}. Let $S$ be an arbitrary set of worlds, and let $w$
be some world (that may or may nor be in $S$). It will be useful to have an
expression that picks out the most similar worlds to $w$, among all the worlds in
$S$. This expression is $\mathrm{Min}^{\prec_w}(S)$, which we have defined as follows in
section \ref{sec:oblig-circ}:
\[
  \mathrm{Min}^{\prec_w}(S) =_\text{def} \{ v: v \in S \land \neg\exists u (u \in S \land u \prec_w v) \}.
\]

Now $\{ u : M,u\models A \}$ is the set of worlds (in model $M$) at which $A$ is
true. So $\mathrm{Min}^{\prec_w}(\{ u : M,u\models A \})$ is the set of those
$A$-worlds that are closest to $w$. We want $A \boxright B$ to be true at $w$
iff $B$ is true at the closest $A$-worlds to $w$.

\begin{definition}{Similarity semantics for $\boxright$}{similaritysemantics}
  If $M$ is a similarity model and $w$ a world in $M$, then\\[1mm]
  $M,w \models A \boxright B$ iff $M,v \models B$ for all
  $v$ in $\mathrm{Min}^{\prec_w}(\{ u: M,u \models A \})$.
\end{definition}

You may notice that $A \boxright B$ works almost exactly like $\Ob(B/A)$ from
section \ref{sec:oblig-circ}. There, I said that for any world $w$ in any deontic
ordering model $M$,

\medskip
\quad$M,w \models \Ob (B/A) \text{ iff } M,v \models B\text{ for all $v$ in $\mathrm{Min}^{\prec_w}(\{ u: wRu $ and $M,u\models A \})$}$.

\medskip \noindent%
The main difference is that conditional obligation is sensitive to an
accessibility relation. If that relation is an equivalence relation then this
makes no difference to the logic.

% Does quasi-connectedness make a difference to the logic? I haven't assumed it
% for deontic ordering models!

Of course, the order $\prec$ in deontic ordering models is supposed to represent
degree of conformity to norms, while the order $\prec$ in similarity models
represents a certain similarity ranking in the evaluation of subjunctive
conditionals. A different type of ordering might be in play when we evaluate
indicative conditionals, which some have argued should also be interpreted as
variably strict. But again, these differences in interpretation don't affect the
logic.

Suppose we add the $\boxright$ operator to the language of standard
propositional logic. The set of sentences in this language that are true at all
worlds in all similarity models is known as \textbf{system V}. There are tree
rules and axiomatic calculi for this system, but they aren't very user-friendly.
We will only explore the system semantically.

To begin, we can check whether \emph{modus ponens} is valid for $\boxright$. That is,
we check whether the truth of $A$ and $A \boxright B$ at a world in a
similarity model entails the truth of $B$.

Assume that $A$ and $A \boxright B$ are true at a world $w$. By definition
\ref{def:similaritysemantics}, the latter means that $B$ is true at all the
closest $A$-worlds to $w$ (at all worlds in
$\mathrm{Min}^{\prec_w}(\{u: M,u\models A\})$). The world $w$ itself is an
$A$-world. If we could show that $w$ is among the closest $A$-worlds to itself
then we could infer that $A$ is true at $w$.

Without further assumptions, however, we can't show this. If we want to validate
\emph{modus ponens}, we must add a further constraint on our models: that every
world is among the closest worlds to itself. More precisely,
\[
  \text{for all worlds $w$ and $v$, $v \nprec_{w} w$.}
\]
This assumption is known as \textbf{Weak Centring}.
% It means that the closeness spheres associated with a world are centred on that
% world: every world is in the sphere of closest worlds around itself.
The logic we get if we impose this constraint is \textbf{system VC}.

\begin{exercise}
  Should we accept Weak Centring for deontic ordering models?
\end{exercise}
\begin{solution}
  No. We don't want $A$ and $\Ob(B/A)$ to entail $B$. Semantically, we don't
  want to assume that every world is among the best worlds relative to its own
  norms.
\end{solution}

\begin{exercise}
  Explain why $A \boxright B$ entails $A \to B$, assuming Weak Centring.
\end{exercise}
\begin{solution}
  Suppose $A \boxright B$ is true at some world $w$ in some model $M$. So $B$ is
  true at all the closest $A$-worlds to $w$. Now either $A$ is true at $w$ or
  $A$ is false at $w$. If $A$ is false at $w$, then $A\to B$ is true at $w$. If
  $A$ is true at $w$, then $w$ is one of the closest $A$-worlds to $w$, by Weak
  Centring; so $B$ is true at $w$; and so $A\to B$ is true at $w$. Either way,
  then, $A\to B$ is true at $w$.
 \end{solution}

\begin{exercise}
  Show that if $A$ is true at no worlds, then $A \boxright B$ is true.
  % This shows how we could define box A.
\end{exercise}
\begin{solution}
  If $A$ is true at no worlds, then $\mathrm{Min}^{\prec_w}(\{u: M,u\models A\})$
  is the empty set. So it is vacuously true that $M,v \models B$ for all
  $v \in \mathrm{Min}^{\prec_w}(\{ u: M,u \models A \})$.
\end{solution}

None of the problematic inferences (E1)--(E5) are valid if the relevant
conditionals are interpreted as variably strict. (E5), for example, would assume
that $p \boxright r $ entails $(p \land q) \boxright r$. But it does not.
We can
give a countermodel with two worlds
\begin{wrapfigure}{r}{5cm}
  \vspace{-5mm}
  \quad
  \begin{tikzpicture}[modal, world/.append style={minimum size=0.5cm}]
    \node[world] (w) [label=above:{$w$}] {$p,r$};
    \node[world] (v) [label=above:{$v$}, right=7mm of w] {$p,q$};
    \node[circle, draw=gray, minimum size=20mm] at (w) (c) {};
    \node[circle, draw=gray, minimum size=50mm] at (w) (c) {};
    % \node[circle, draw=gray, minimum size=20mm] at (v) (c) {};
  \end{tikzpicture}
  \vspace{-10mm}
\end{wrapfigure}
$w$ and $v$; $p$ is true at both worlds, $q$
is true only at $v$, and $r$ only at $w$; if $w$ is closer
to itself than $v$,
then $p \boxright r$ is true at $w$ (because the closest $p$-worlds to $w$ are
all $r$-worlds), but $(p \land q) \boxright r$ is false at $w$ (because the
closest $(p\land q)$-worlds to $w$ aren't all $r$-worlds).

The diagram on the right represents this model. The circles around $w$ depict
the similarity spheres. $w$ is closer to $w$ than $v$ because it is in the
innermost sphere around $w$, while $v$ is only in the second sphere. (If $v$
were also in the innermost sphere then the two worlds would be equally close to
$w$. That's allowed.) In general, we can represent the assumption that a world $v$
is closer to a world $w$ than a world $u$ ($v \prec_w u$) by putting $v$ is in a
closer sphere around $w$ than $u$. I have not drawn any spheres around $v$
because it doesn't matter what these look like.

% Can we turn this approach into a pictorial proof method? 1. start by putting
% the target sentence(s) into a world w. 2. Draw a sphere around w. 3. Expand
% each world as a non-modal tree. 3a. Rule for A []-> B at w: put A->B in all
% worlds outwards from w until and including worlds in spheres within which A is
% true somewhere. 3b. Rule for not(A []-> B): add a world v with A and -B, and
% make it one of the closest A-worlds. I.e., if A at w, then add v to innermost
% sphere. If not-A at w, draw a sphere around just w and put v outside it. ....?

\begin{exercise}
  Draw countermodels showing that (E1)--(E4) are invalid if the conditionals are
  translated as statements of the form $A \boxright B$. (Hint: You never need
  more than two worlds.)
\end{exercise}
\begin{solution}
    (E1) is an inference from $q$ to $p \boxright q$. To show that this is invalid, we need to give a model in which $q$ is true at some world ($w$) while $p \boxright q$ is false (at $w$). 
    
  \begin{tikzpicture}[modal, world/.append style={minimum size=0.5cm}]
    \node[world] (w) [label=above:{$w$}] {$q$};
    \node[world] (v) [label=above:{$v$}, right=7mm of w] {$p$};
    \node[circle, draw=gray, minimum size=20mm] at (w) (c) {};
    \node[circle, draw=gray, minimum size=40mm] at (w) (c) {};
  \end{tikzpicture}
  
  This model also shows that (E2) and (E3) are invalid.
  (E2) is an inference from $\neg p$ to $p \boxright q$. In the model,
  $\neg p$ is true at $w$ but $p \boxright q$ is false.
  (E3) is an inference from $\neg(p \boxright q)$ to $p$. In the model,
  $\neg(p \boxright q)$ is true at $w$ but $p$ is false.

  (E4) is an inference from $p \boxright q$ to $\neg q \boxright \neg p$.
  In the following model, the premise is true at $w$ and the conclusion false.
  
  \begin{tikzpicture}[modal, world/.append style={minimum size=0.5cm}]
    \node[world] (w) [label=above:{$w$}] {$p,q$};
    \node[world] (v) [label=above:{$v$}, right=7mm of w] {$p$};
    \node[circle, draw=gray, minimum size=20mm] at (w) (c) {};
    \node[circle, draw=gray, minimum size=40mm] at (w) (c) {};
  \end{tikzpicture}
\end{solution}

The logic of variably strict conditionals is weaker than the logic of strict
conditionals. Some have argued that it is too weak to explain our
reasoning with conditionals. It is, for example, not hard to see that the
following statements are all false. (The corresponding statements for
$\strictif$ are true; see exercise \ref{ex:sda-import}.)
\begin{enumerate}[leftmargin=10mm]
  \itemsep-1mm
\item $p \boxright q, q \boxright r \models p \boxright r$
\item $((p \lor q) \boxright r) \models (p \boxright r) \land (q \boxright r)$
\item $p \boxright (q \boxright r) \models (p \land q) \boxright r$
\end{enumerate}
If English conditionals are variably strict, this means (for example) that we can't 
infer `if $p$ then $r$' from `if $p$ then $q$' and `if $q$ then $r$'. But isn't
this a valid inference?

Well, perhaps not. Stalnaker gave the following counterexample, using cold-war
era subjunctive conditionals.
\begin{quote}
  If J.\ Edgar Hoover had been born a Russian, he would be a communist.\\
  If Hoover were a communist, he would be a traitor.\\
  Therefore, if Hoover had been born a Russian, he would be a traitor.
\end{quote}

\begin{exercise}
  Can you find a case where `if $p$ or $q$ then $r$' does not appear to entail
  `if $p$ then $r$' and `if $q$ then $r$'? You can use either indicative or
  subjunctive conditionals. (Hint: Try to find a case in which `if $p$ or $q$
  then $p$' sounds acceptable.)
\end{exercise}
\begin{solution}
  Frances has never learnt a foreign language, although she would have loved to
  learn French. If Frances had been given a choice between learning French and
  learning Italian, she would have chosen French. \emph{If Frances had learned
    French or Italian then she would have learned French.} It does not follow that
  if Frances had learned Italian then she would have learned French.

  The same style of example works for indicative conditionals.
\end{solution}

% \begin{exercise}
%   Are \emph{modus tollens} and contraposition valid for $\boxright$? That is,
%   (a) do $A$ and $A \boxright B$ always entail $B$, and (b) does
%   $A \boxright B$ always entail $\neg B \boxright \neg A$?
% \end{exercise}

The semantics I have presented for $\boxright$ is a middle ground between that
of Lewis and Stalnaker. Stalnaker assumes that $\prec_w$ is not just
quasi-connected, but connected: for any $w,v,u$, either $v \prec_w u$ or $v=u$
or $u \prec_w v$. (`$v=u$' means that $v$ and $u$ are the same world.) This
rules out ties in similarity: no sphere contains more than one world.

Stalnaker's logic (called \textbf{C2}) is stronger than Lewis's VC. The
following principle of ``Conditional Excluded Middle'' is C2-valid but not VC-valid:
%
\principle{CEM}{(A \boxright B) \lor (A \boxright \neg B)}

%  NB: Uniqueness alone doesn't entail CEM, only Uniqueness + Limit.

Whether conditionals in natural language satisfy Conditional Excluded Middle is
a matter of ongoing debate. On the one hand, it is natural think that `it is not
the case that if $p$ then $q$' entails `if $p$ then not $q$', which suggests
that the principle is valid. On the other hand, suppose I have a number of coins
in my pocket, none of which I have tossed. What would have happened if I had
tossed one of the coins? Arguably, I might have gotten heads and I might have
gotten tails. Either result is possible, but neither \emph{would} have come
about.

\begin{exercise}
  Explain why the following statements are true, for all $A,B,C$:
  \begin{exlist}
  \item $A \land B \models_{C2} A \boxright B$
  \item $A \boxright (B\lor C) \models_{C2} (A \boxright B) \lor (A \boxright C)$
  \end{exlist}
\end{exercise}
\begin{solution}
  \begin{sollist}
    \item Assume $A\land B$ is true at some world $w$ in some model $M$. By
    Centring, $w$ is among the closest $A$-worlds to $w$. By connectedness, $w$
    is the unique closest $A$-world to $w$. So $B$ is true at all closest
    $A$-worlds to $w$.
    \item Assume $A \boxright (B\lor C)$ is true at some world $w$ in some model
    $M$. So all the closest $A$-worlds to $w$ are $(B\lor C)$-worlds. If there
    are no $A$-worlds then $A \boxright B$ and $A \boxright C$ are both true. If
    there are $A$-worlds then Stalnaker's semantics implies that there is a
    unique closest $A$-world $v$ to $w$. Since $B\lor C$ is true at $v$, either $B$ or $C$ must be true at $v$. So either $B$ is true at all closest $A$-worlds to $w$ or $C$ is true at all closest $A$-worlds to $v$.
  \end{sollist}
\end{solution}

Lewis not only rejects connectedness, but also the Limit Assumption. He argued
that there might be an infinite chain of ever closer $A$-worlds. Definition
\ref{def:similaritysemantics} implies that if there are no closest $A$-worlds
then any sentence of the form $A \boxright B$ is true. That does not seem right.
Lewis therefore gives a more complicated semantics:

\begin{quote}
  $M,w \models A \boxright B$ iff either there is no $v$ for
  which $M,v\models A$ or there is some world $v$ such that
  $M,v\models A$ and for all $u \prec_w v$, $M,w \models A \to B$.
\end{quote}
%
It turns out that it makes no difference to the logic whether we impose the
Limit Assumption and use the old definition or don't impose the Limit Assumption
and use Lewis's new definition. The same sentences are valid either
way. % (Lewis 1973, p.444).

% Rini and Cresswell 58 use a 4-place relation C to evaluate >, where for any
% time t and worlds w1, w2, w3, C(t,w1,w2,w3) iff at t, w2 is closer to w1 than
% w3. I can probably fold the t into w1. RC point out that Aqvist 1973 also
% invented this kind of semantics.

\iffalse

Let's say that for any sentence $A$, a world $v$ is \emph{$A$-accessible} from
$w$ (for short, $wR_Av$) iff $v$ is one of the closest $A$-worlds to $w$; that
is, iff $v \in \mathrm{Min}^{\prec_w}(\{u: M,u \models A\})$. Definition
\ref{def:similaritysemantics} then states that $A \boxright B$ is true at a
world $w$ iff $B$ is true at all worlds $A$-accessible from $w$. These are the
standard truth-conditions for the box, except that accessibility is relativised
to the antecedent $A$.

We can therefore adapt the standard tree rules for the box to reason with
$\boxright$, as follows.

\bigskip
\begin{center}
\begin{minipage}{0.4\textwidth} \centering
\tree{
  \nnode{18}{}{$A \boxright B$}{\omega}{}\\
  \dotbelownode{18}{}{$\omega R_A v$}{}{}\\
  \\
  \nnode{18}{}{$B$}{\nu}{}\\
  \Kk[18]{0}{\color{red}$\uparrow$}\\
  \Kk[18]{0}{\color{red}\small old}
}
\end{minipage}
\begin{minipage}{0.4\textwidth}\centering
\tree{
  \dotbelownode{18}{}{$\neg (A \boxright B)$}{\omega}{}\\
  \\
  \nnode{18}{}{$\omega R_A v$}{}{}\\
  \nnode{18}{}{$\neg B$}{\nu}{}\\
  \Kk[18]{0}{\color{red}$\uparrow$}\\
  \Kk[18]{0}{\color{red}\small new}
}
\end{minipage}
\end{center}
\bigskip

To get a complete tree system, we need further rules. For example, the above two
rules don't account for the fact that the closest $A$-worlds are always
$A$-worlds. They also don't account for the fact that every $A$-world is among
the closest $A$-worlds to itself (by weak centring). We can add two more rules
to fill these gaps.

\bigskip
\begin{center}
  \begin{minipage}[t]{0.4\textwidth} \centering
    \hspace{-10mm}Truth:
    \medskip
    
\tree{
  \dotbelownode{12}{}{$\omega R_A \nu$}{}{}\\
  \\
  \nnode{12}{}{$A$}{\nu}{}\\
}
\end{minipage}
\begin{minipage}[t]{0.4\textwidth} \centering
  \hspace{-10mm}Centring:
    \medskip
  
\tree{
  \dotbelownode{12}{}{$A$}{\omega}{}\\
  \\
  \nnode{12}{}{$\omega R_A \omega$}{}{}\\
}
\end{minipage}
\end{center}
\bigskip

The resulting tree system is still not complete, because it doesn't reflect
interactions between different accessibility relations. For example, if the
worlds that are $A$-accessible from some world $w$ include $B$-worlds, then the
$A\land B$-accessible worlds from $w$ must be contained within the worlds
$A$-accessible from $w$. The complete tree rules for the $\boxright$ operator
turn out to be rather complicated. I will leave it at the above four rules,
which suffice to establish many useful facts about variably strict conditionals.

% Priest points out that for Lewis-Stalnaker we have more conditions than in the
% basic "plural selection" semantics that underlines his tableaux. In particular,
% we have: (3) if $A$ is consistent, then there is always a non-empty set of
% closest $A$-worlds; (4) if the closest $A$-worlds are a subset of the B worlds,
% and the closest B worlds are a subset of the A-worlds, then the closest A-worlds
% area the clsoest B-worlds; (5) if there are some B-worlds among the clsoest
% A-worlds, then the closest A-and-B worlds are among the closest A-wrolds. [p.92]
% This means that e.g. $A > B, B> A \models (A > C) \leftrightarrow (B > C)$ is
% valid in LS, but not provable with a tableaux. "There are presently no known
% tableau systems of the kind used in this book for S [Lewis/Stalnaker] (and its
% extensions that we will meet in the next section)". p.93. So I might mention
% that the rules are incomplete, meaning that anything that can be proved is
% indeed valid, but that caution is needed when concluding that something is
% invalid.

% Zach 2018 points out that the logic and semantics Priest discusses is due to
% Chellas; the tableau rules are Priest's. ``We give additional branch extension
% rules for Priest’s system which result in sound and complete tableau systems for
% CK and [Lewis's] VC. These systems are, however, non-analytic in that the cut
% rule is included. ... These are not the first tableau systems for Lewis’s VC .
% Gent [1992], based on work of de Swart [1983], has given a tableau system for VC
% . More recently, Negri and Sbardolini [2016] have offered a cut-free complete
% sequ ent calculus for VC . These approaches are all based on Lewis’s semantics
% based on rela tive proximity of worlds and incorporate the ordering relation
% between w orlds into the syntax.''

Here is a tree proof to show (once more) that \emph{modus ponens} is valid.

\begin{center}
  \tree[3]{
    & \nnode{18}{1.}{$A \boxright B$}{w}{(Ass.)} & \\
    & \nnode{18}{2.}{$A$}{w}{(Ass.)} & \\
    & \nnode{18}{3.}{$\neg B$}{w}{(Ass.)} & \\
    & \nnode{18}{4.}{$wR_Aw$}{}{\quad\quad(Centring)} & \\
    & \nnodeclosed{18}{5.}{$B$}{w}{(1,4)} & 
  }
\end{center}

\begin{exercise}
  Give tree proofs for the following statements.
  \begin{exlist}
  \item $A, \neg B \models \neg(A\boxright B)$ 
  \item $A\boxright B, A\boxright C \models  A \boxright (B \land C)$ % from Girle
  \item $A \boxright (B \land C) \models (A \boxright B) \land (A\boxright C)$
  \item $A\boxright \neg A \models A\boxright B$ 
  \end{exlist}
\end{exercise}

Since the tree rules I have presented are sound, you can be sure that whenever a
tree closes then the tested entailment or validity holds.  But since the rules
are not complete, care is required when a tree doesn't close. You always need to
check if a model read off from an open tree is an actual countermodel.

Constructing countermodels from open trees is at any rate not entirely
straightforward. By way of illustration, let's show that
$A\boxright B$ and $B\boxright C$ does not entail $A \boxright C$. The
tree starts like this.

\begin{center}
  \tree[3]{
    & \nnode{20}{1.}{$A \boxright B$}{w}{(Ass.)} & \\
    & \nnode{20}{2.}{$B \boxright C$}{w}{(Ass.)} & \\
    & \nnodeticked{20}{3.}{$\neg(A \boxright C)$}{w}{(Ass.)} & \\
    & \nnode{20}{4.}{$wR_Av$}{}{(3)} & \\
    & \nnode{20}{5.}{$\neg C$}{v}{(3)} & \\
    & \nnode{20}{6.}{$B$}{v}{(1,4)} & \\
    & \nnode{20}{7.}{$A$}{v}{\quad\;(4,Truth)} & 
  }
\end{center}
The Centring rule would allow us to add six more lines, but they
wouldn't be useful, so let's stop here.
% & \nnode{18}{2.}{$wR_{\neg C}w$}{}{(5,Centring)} & \\

The open tree suggests that there is a countermodel with two worlds, $w$ and
$v$. At $v$, $A$ and $B$ are true. We also have $wR_Av$, so $v$ is among the
closest $A$-worlds to $w$. Since $wR_A w$ is not in the tree, let's assume that
$w$ is not among the closest $A$-worlds to $w$, which also means that $A$ is
false at $w$. We don't have $wR_B v$ on the tree either. So even though $B$ is
true at $v$, $v$ is not among the closest $B$-worlds to $w$. We can ensure this
by assuming that $B$ is true at $w$ itself, and that $w$ is the unique closest
$B$-world from itself. Now you can verify that $A \boxright B$ and
$B \boxright C$ are both true at $w$ while $A \boxright C$ is false.

% Priest gives some rules for reading off countermodels: For the initial logic:
% ``Counter-models are read off from the tableau in a natural way. If there is
% something of the form A > B or ¬(A > B) on the branch, then RA is as the
% information about rA on the branch specifies. Otherwise, RA may be arbitrary.''
% For the full logic: ``If A does not occur as the antecedent of a conditional or
% negated conditional at a node, we can no longer allow RA to be arbitrary,
% however, since it must satisfy (1) [centring] and (2) [A is true at any
% R(A)-accessible world]. The simplest trick is to let [the set of closest
% A-worlds to w] = [A] [i.e., the set of all A-worlds] (for every w). With this
% definition, (1) and (2) are clearly satisfied.'' He adds a footnote: ``This is
% legitimate, since [the set of closest A-worlds] is not required to define the
% truth value of A at a world. To evaluate the truth value of A at a world, one
% needs to know only [the set of closest B-worlds] for those B that occur as the
% antecedents of conditionals within A.'' I don't get it. Doesn't this treat the
% conditional as strict? Ah. $\models_S \Phi$ implies $\models_C \Phi$; i.e. if
% something is valid for strict implication then it's also valid for variably
% strict implication. (Right? Might be a good exercise.) So $\not\models_C \Phi$
% implies $\not\models_S\Phi$. So any invalidity in $C$ is also an invalidity in
% $S$. And so (??) any countermodel in $C$ is also a countermodel in $S$.

\fi


\section{Restrictors}
\label{sec:restrictor-analysis}

Consider these two statements.
\begin{enumerate}[leftmargin=10mm]
  \itemsep-1mm
  \item[(1)] If it rains we always stay inside.
  \item[(2)] If it rains we sometimes stay inside.
\end{enumerate}
On its most natural reading, (1) says that we stay inside at all times at which
it rains. We can express this in $\L_{M}$, using the box as a universal
quantifier over the relevant times. (So $\Box A$ now means `always $A$'.) The
translation would be $\Box(r \to s)$.

One might expect that (2) should then be translated as $\Diamond(r \to s)$,
where the diamond is an existential quantifier over the relevant times
(`sometimes'). But $\Diamond(r \to s)$ is equivalent to
$\Diamond(\neg r \lor s)$. This is true whenever $\Diamond \neg r$ is true. (2),
however, isn't true simply because it doesn't always rain. On its most salient
reading, (2) says there are times at which it rains \emph{and} we stay inside.
Its correct translation is $\Diamond(r \land s)$.

This is a little surprising, given that (2) seems to contain a conditional. Does
the conditional here express a conjunction?

Things get worse if we look at (3).
\begin{enumerate}[leftmargin=10mm]
  \item[(3)] If it rains we usually stay inside.
\end{enumerate}
Let's introduce an operator $\Mostly$ for `usually', so that $\Mostly A$ is true
at a time iff $A$ is true at \emph{most} times. Can you translate (3)
with the help of $\Mostly$?

You can't. Neither $\Mostly(r \to s)$ nor $\Mostly(r \land s)$ capture the
intended meaning of (3). $\Mostly(r \land s)$ entails that $r$ is usually true.
But (3) doesn't entail that it usually rains. $\Mostly(r \to s)$ is true as long
as $r$ is usually false, even if we're always outside when it is raining.
You could try to bring in some of the new kinds of conditional that we've
encountered in the previous sections. How about $\Mostly(r \boxright s)$, or
$\Mostly(r \strictif s)$, or $r \boxright \Mostly s$, or
$r \strictif \Mostly s$? None of these are adequate.

The problem is that (3) doesn't say, of any particular proposition, that it is
true at most times. It doesn't say that among all times, most are such-and-such.
Rather, it says that \emph{among times at which it rains}, most times are times
at which we stay inside. The function of the `if'-clause in (3) is to
\textbf{restrict the domain} of times over which the `usually' operator
quantifies.

% We can formalize (3) if we treat $\Mostly$ as a binary operator, taking two
% propositions as arguments: $\Mostly(s/r)$. The semantics for the two forms of
% $\Mostly$ would look as follows, assuming that we are quantifying over times.

% \bigskip
% \begin{tabular}{ll}
%   $M,t \models \Mostly A$ &iff $M,s \models A$ for most times $s$.\\
%   $M,t \models \Mostly(A/B)$ &iff $M,s \models A$ for most times $s$ such that
%                                $M,s \models B$.
% \end{tabular}

Now return to (1) and (2). Suppose that here, too, the `if'-clause serves to
restrict the domain of times, so that `always' and `sometimes' only quantify
over times at which it rains. On that hypothesis, (1) says that \emph{among
  times at which it rains}, all times are times at which we stay inside, and (2)
says that \emph{among times at which it rains}, some times are times at which we
stay inside. This is indeed what (1) and (2) mean, on their most salient
interpretation.

As it turns out, `among $r$-times, all times are $s$-times' is equivalent to `all
times are not-$r$-times or $s$-times'. That's why we can formalize (1) as
$\Box(r \to s)$. `Among $r$-times, some times are $s$-times', on the other hand,
is equivalent to `some times are $r$-times and $s$-times'. That's why we can
formalize (2) as $\Diamond(r \land s)$. It would be wrong to think that the
conditional in (1) is material, the conditional in (2) is a conjunction, and the
conditional in (3) is something else altogether. A much better explanation is
that the `if'-clause in (1) does the exact same thing as in (2) and (3). In each
case, it restricts the domain of times over which the relevant operators
quantify.

We can arguably see the same effect in (4) and (5).
\begin{enumerate}[leftmargin=10mm]
  \itemsep-1mm  
  \item[(4)] If the lights are on, Ada must be in her office.
  \item[(5)] If the lights are on, Ada might be in her office.
\end{enumerate}
Letting the box express epistemic necessity, we can translate (4) as
$\Box(p \to q)$. But (5) can't be translated as $\Diamond(p \to q)$, which would
be equivalent to $\Diamond(\neg p \lor q)$. Nor can we translate (5) as
$p \to \Diamond q$, which is entailed by $\Diamond q$. It is easy to think of
scenarios in which (5) is false even though `Ada might be in her office' is
true. The correct translation of (5) is plausibly $\Diamond(p \land q)$. The
sentence is true iff there is an epistemically accessible world at which the
lights are on and Ada is in her office.

As before, we can understand what is going if we assume that the `if'-clause in
(4) and (5) functions as a restrictor. The `if'-clause restricts the domain of
worlds over which `must' and `might' quantify. (4) says that \emph{among
  epistemically possible worlds at which the lights are on}, all worlds are
worlds at which Ada is in her office. (5) says that \emph{among epistemically
  possible worlds at which the lights are on}, some worlds are worlds at which
Ada is in her office.

\begin{exercise}
  Translate `all dogs are barking' and `some dogs are barking' into the language
  of predicate logic. Can you translate `most dogs are barking' if you add a
  `most' quantifier $\Mostly$ so that $\Mostly x Fx$ is true iff most things
  satisfy $Fx$?
\end{exercise}
\begin{solution}
  `All dogs are barking': $\forall x(Dx \to Bx)$\\
  `Some dogs are barking': $\exists x(Dx \land Bx)$\\
  `Most dogs are barking' cannot be translated in terms of $\Mostly x$. We need
  a binary quantifier: $\Mostly x(Bx / Dx)$
  % (There is a strong parallel here to explicit (nominal) quantifiers in
  % natural language. We usually translate `all $F$ are $G$' as
  % $\forall x (Fx \to Gx)$, and `some $F$ are $G$' as
  % $\exists x (Fx \land Gx)$. This works, but it is a bit strange that what
  % appear to be entirely analogous constructions in English get such different
  % translations. Moreover, the approach breaks down for `most'. `Most $F$ are
  % $G$' cannot be analysed in terms of a quantifier $Mx$ that applies to
  % individual open sentences. In \emph{generalised quantifier theory}, all
  % quantifiers are instead treated as two-place functions, so that we can write
  % $\forall x (Gx/Fx)$, $\exists x (Gx/Fx)$, and $Mx (Gx/Fx)$. (Actually, these
  % are more commonly written $[\forall x: Fx]Gx$, \ldots.) Semantically, the
  % extra argument place selects the domain in which it is claimed that
  % all/some/most things are $G$.)
\end{solution}

% ``The data involving modification by only and even, and VP ellipsis phenomena
% provide strong evidence against the view that the antecedent and consequent of
% conditionals are coordinated. These data support the view that if-clauses are
% adverbials, like temporal phrases and clauses. Furthermore, pronominalization by
% then suggests that if-clauses are advervials, since their anaphoric reflex -
% then - is an adverb''. Bhatt and Pancheva 2006

The hypothesis that `if'-clauses are restrictors also sheds light on the problem
of conditional obligation.
\begin{enumerate}[leftmargin=10mm]
  \itemsep-1mm  
  \item[(6)] Jones ought to help his neighbours.
  \item[(7)] If Jones doesn't help his neighbours, he ought to not tell them that he's coming.
\end{enumerate}
In chapter \ref{ch:deontic}, we analyzed `ought' as a quantifier over the best
of the circumstantially accessible worlds. On this approach, (6) says that among
the accessible worlds, all the best ones are worlds at which Jones helps his
neighbours. Suppose the `if'-clause in (7) serves to restrict the domain of
worlds, excluding worlds at which Jones helps his neighbours. We then predict
(7) to state that \emph{among the accessible worlds at which Jones doesn't help
  his neighbours}, all the best worlds are worlds at which Jones doesn't tell
his neighbours that he's coming. This can't be expressed by combining the
monadic $\Ob$ quantifier with truth-functional connectives. Hence we had to
introduce a primitive binary operator $\Ob(\cdot/\cdot)$.

% As Richmond Thomason argued in 1981, A proper theory of conditional
% obligation \ldots will be the product of two separate components: a
% theory of the conditional and a theory of obligation.

The upshot of all this is that we can make sense of a wide range of puzzling
phenomena by assuming that `if'-clauses are restrictors. Their function is to
restrict the domain or worlds or times over which modal operators quantify.

What, then, is the purpose of `if'-clauses in ``bare'' conditionals like (8) and
(9), where there are no modal operators to restrict?
\begin{enumerate}[leftmargin=10mm]
  \itemsep-1mm
  \item[(8)] If Shakespeare didn't write \emph{Hamlet}, then someone else did.
  \item[(9)] If Shakespeare hadn't written \emph{Hamlet}, then someone else
        would have.
\end{enumerate}

Here opinions vary. One possibility, prominently defended by the linguist
Angelika Kratzer, is that even bare conditionals contain modal operators.
Arguably, `would' in (9) functions as a kind of box. If this box is a simple
quantifier over circumstantially accessible worlds, and the `if'-clause in (9)
restricts its domain, then (9) can be formalized as $\Box(p \to q)$. If, on the
other hand, `would' in (9) works more light `ought' -- if it quantifyies over the
\emph{closest} of the accessible worlds --, and the `if'-clause restricts the
domain of accessible worlds, then the resulting truth-conditions are those of
$p \boxright q$. Both the strict and the variably strict analysis of (9) are
therefore compatible with the hypothesis that `if'-clauses are restrictors.

What about (8)? This sentence really doesn't appear to contain a relevant modal.
Kratzer suggests that it contains an unpronounced epistemic `must': (8) says
that if Shakespeare didn't write \emph{Hamlet} then someone else \emph{must}
have written \emph{Hamlet}. Assuming that the `if'-clause restricts the domain
of this operator, bare indicative conditionals would be equivalent to
strict epistemic conditionals.

\begin{exercise}
  Suppose bare indicative conditionals like (8) contain a box operator $\Box$
  whose accessibility relation relates each world to itself and to no other
  world. (This is a redundant operator insofar as $\Box A$ is equivalent to $A$.)
  Assume the `if'-clause restricts the domain of that operator. What are the
  resulting truth-conditions of (8)?
\end{exercise}
\begin{solution}
  On this proposal, bare indicative conditionals like (8) are material
  conditionals. If $p$ is true and $q$ is false then there is an accessible
  $p$-world at which $q$ is false, and so $q$ is not true at all accessible
  worlds at which $p$ is true. In all other cases, $q$ \emph{is} true at all
  accessible worlds at which $p$ is true.
\end{solution}

\begin{exercise}
  Besides ``would counterfactuals'' there are also ``might counterfactuals'' like
  \begin{enumerate}[leftmargin=10mm]
    \item[(10)] If I had played the lottery, I might have won.
  \end{enumerate}
  Suppose `might' is the dual of `would', and suppose the `if'-clause in (10)
  restricts the domain of worlds over which `might' quantifies. It follows that
  `if $A$ then might $B$' is true iff $B$ holds at some of the
  closest/accessible $A$-worlds. (`Closest' or `accessible' depending on how we
  understand the `would'/`might' operators.) Can you see why this casts doubt on
  the validity of Conditional Excluded Middle?
\end{exercise}
\begin{solution}
  Conditional Excluded Middle is valid iff there is never more than one
  closest/accessible $A$-world. On that assumption, `some closest/accessible
  $A$-world is a $B$-world' entails `all closest/accessible $A$-worlds are
  $B$-worlds'. But (10) does not entail `If I had played the lottery, I would
  have won'.
\end{solution}


\iffalse

We don't have a versatile restrictor device in $\L_{M}$. For many purposes it
isn't needed. We can use $\Box(A \to B)$ or $\Diamond(A \land B)$ to express
restricted necessity or possibility claims. Could we also add a conditional
operator $>$ that serves as a general restrictor?

Informally, we want $A > B$ to be true at a world $w$ in a model $M$ iff $B$ is
true at $w$ in a model $M'$ that is like $M$ except that all not-$A$ worlds are
removed. To achieve this, let's make truth relative to an added parameter $U$,
the ``universe of live possibilities''. The truth-functional connectives behave
in the usual way. We have the following clauses for sentence letters, the box,
and the restrictor conditional.

\begin{definition}{Restrictor semantics for $>$}{restrictorsemantics}
  If $\Mfr = \t{W,R,V}$ is a Kripke model, $w$ is a member of $W$, $A$ is a
  sentence letter, and $B,C$ are $\L_M$-sentences, then \medskip
  \begin{tabular}{lll}
    (a) & $M,U,w \models A$ &iff $w$ is in $V(A)$.\\
    (b) & $M,U,w \models \neg B$ &iff $M,U,w\not\models B$.\\
    (c) & $M,U,w \models A \to B$ &iff $M,U,w\not\models B$ or $M,U,w \models C.\\
    % $M,U,w \models A$ iff $M,U,w \models A$ and $w \in V(A)$;\\
    % without this, 'if p q' is equivalent to 'q'.
    (d) & $M,U,w \models \Box B$ &iff $M,v \models B$ for all $v$ in $U$ such that $wRv$.\\
    (e) & $M,U,w \models B > C$ &iff $M,U,w \not\models B$ or $M,\{ v \in U: M,U,v \models B \},w \models C$.
    % "At all worlds, blanking out the A worlds, B" should mean the same as "at all A-worlds, B". So "blanking out the A worlds, B" must be true at w iff w is either a not-A world or a B-world. I.e., w |= A>B iff w |= A->B ?! We say w |= A>B iff [A],w |= B. This should be vacuously true if w is not in [A]. Can we ensure in the compositional semantics that U,w occurs on the LHS of |= only if w \in U? Sure. The condition is trivially satisfied for U=W. So we only need to adjust the clause for >: instead of saying that U,w |= A>B iff U \cap [A],w |= B, we say that U,w |= A>B iff either w \not\in [A] or U \cap [A],w |= B. Equivalently, U,w |= A>B iff U \cap [A],w |= A->B. It might be better to treat w \in U as a presupposition. -- All of that is problematic if we want A > []B to be non-vacuous if A is false! "Blanking out non-A worlds, all worlds are B-worlds" should not be vacuous.
  \end{tabular}
\end{definition}
%
Clause (d) says that $\Box
B$ is true at $w$ iff all accessible worlds among the live possibilities $U$ are $B$-worlds. Clause (e) says that $B
>
C$ is true at $w$ iff restricting the live possibilities $U$ to $B$-worlds renders $C$ true at $w$, provided $w$ is itself a $B$-world; if it is not, then $B
> C$ counts as vacuously true at $w$. 

We define truth relative to a world and a model by stipulating that for any sentence $A$, $M,w
\models A$ iff $M,W,w \models A$.

It is easy to see that $p > \Box q$ is equivalent to $\Box(p \to q)$, in the
sense that $M,w \models p > \Box q$ iff $M,w \models \Box(p \to q)$. Both
sentences are true at a world $w$ iff $q$ is true at all $w$-accessible worlds
at which $p$ is true.
% check limit case: if there's no (accessible) p-world then p > []A is true iff
% A is true at all v in the empty set; so p > []A is vacuously true, just like
% [](p->q).
In the same way, $p > \Diamond q$ is equivalent to $\Diamond(p \land q)$.
% limit case: if there's no (accessible) p-world then p > <>A is true iff A is
% true at some v in the empty set, which is never the case.

If we add the restrictor $>$ to an ordering model, where $\Box A$ quantifies
over the minimal $U$-worlds (or the minimal accessible $U$-worlds), $p > \Box q$
becomes equivalent to $p \boxright q$. In deontic ordering models, this, in
turn, is equivalent to $\Ob(q/p)$.

% What happens in A > <><>B? The restrictor account suggests that this means
% <>(A & <>B). My semantics makes it equivalent to <><>(A & B).

% Covic and Egre discuss a model in which $R$ is shifted: $w,R \models \Box(A/B)$
% iff $w,R' \models \Box A$ where $R'(w)$ is $R(w)$ restricted by the truth-set of
% $B$ relative to $R,\preceq$. Seems easier with modal bases: the if-clause is
% added to the modal base.

\fi

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