by George N Schlesinger, December 15, 1982

In an important paper, ‘Is epistemic logic possible’, Max 0. Hocutt has

raised the question whether epistemic logic consists of logical truths in which

epistemic terms occur essentially. Admittedly there are theorems which

follow simply from the definition of ‘a knows that p’. However, this does

not amount to a unique feature of the concept of knowledge that would

explain why a specialogic should be constructed around it.

I shall try to show that epistemic logic is definitely ‘possible’ since there

are unquestionably statements expressing logical truths in which epistemic

terms occur essentially. On flrst thought it may appear that

(a) (p -+ q) & JBsp – JBsq

is a good example of such a statement. Unlike Bsq the question whether

JBsq holds does not seem to depend on s’s logical acumen or state of mind.

Even if s fails actually to grasp the connection between p and q, since objectively

speaking q is ‘included’ logically in p which s is justifled in believing,

there would seem no reason to deny (a).

Admittedly (a) has been denied recently, but I believe, on the basis of

faulty argument. Steven Levy in a paper entitled ‘Do you know everything

you know?’ questions the validity of

(b) Ksp & (p – q) – RBsq

where ‘RBsq’ stands for ‘it is rational for s to believe that q’. He says:

Suppose that s knows that p, that p entails q and that s correctly infers q from p. We

may, perhaps, be tempted to approve that any belief formed by s that q, as a result of

this process is a rational belief without inquiring further into the matter. But to do

so would be a mistake. Suppose that s also has a firm belief that r (it makes no difference

whether or not this belief is rational). Suppose further, that r entails not -q and that s

correctly infers not -q from r. In such a situation it would hardly be rational of s to

believe that q. Rather, s should be required to escape the dilemma either by further

inquiry or by suspending any belief that q or not -q. Thus, even though the antecedent

of (a) is satisfied, given s’s other beliefs it would not be rational for him to believe q.^{2}

This argument is erroneous. After all there are two possibilities:

(1) r is not sufficiently well established. Thus in the view of the fact that

p, which is known by s to entail q is strongly enough supported, s is in fact

not justified in holding r (which he has correctly inferred as being inconsistent

with q).

(2) s is objectively justified in continuing to subscribe to r since r has been

firrmly enough established.

Now in case (1), s either adopts the attitude that is required by the standards

of rationality or he does not. Should he choose the first alternative

there is certainly no problem; s in fact rejects r and can therefore hold q without

any reservations. Undoubtedly therefore we may insist that the consequent

(b) is as true as its antecedent. Should he adopt the second attitude,

(b) still remains true since how the antecedent, and not only the consequent,

is false! Given that s believes r and thus -q and assuming that he knows that

-q

– ,p, it follows that -Bsq and thus -Bsp.

In case (2) of course it follows that q has to be withdrawn because of its

incompatibility with the well established r. Thus it can no longer be the case

that RBsq nor that RBsp and therefore -Ksp. Once more therefore (b)

remains true since its antecedent is false.

To summarize: Levy’s example is harmless, since no matter what the

circumstances are, either RBsq remains true, or if not, Ksp is false too.

However, (a) may be rejected on other, legitimate grounds. Many philosophers

hold that q is not justified in believing that p even if p can be established

on the basis of what s knows, as long as a himself is not aware of the

way this is to be done. Suppose that Goldbach’s conjecture is true and I

believe this to be so and there exist a rigorous proof showing this to be so

however I am ignorant of this proof and have not been assured of the truth

of Goldbach’s conjecture by any expert who is familiar with the proof, then

I can neither be said to know that Goldbach’s conjecture is true nor that I

am justified in believing this to be so. In other words, the question whether

‘JBsp. is true or false is not determined solely by such objective factors as the

evidence s possesses and the logical relationship between it and p. Admittedly

it is a necessary condition that there be adequate evidence but unless s also

realized this, ‘JBsp’ is not true.

Let me therefore introduce the symbol ‘JB*sp’ to mean ‘objectively speaking

p can be established on the basis of information in s’s possession’ which of

course amounts to less than ‘JBsp’ that stands for ‘s is actually justifiled in

believing that p’. It may seem then that

(c) (p +q)&JBsp ?JB*sq

represents a logical truth. It turns out however that even (c) will not do.

Suppose that s has adequate evidence that p is true and thus JBsp even though

in fact p is false. Also suppose that s is incapable of inferring q from p but can

clearly see that q is false. In this case JB*sq is false in spite of the fact that

the antecedent of (c) is true.^{3}

However (c) may be amended in a number of ways to yield a valid expression.

It is possible for instance to substitute ‘K’ for ‘JB’ and obtain

(d1) (pt q) & Ksp – JB *sq

The previous objection does not apply to (dl) since no longer can p be false

given that s knows that p. Now (dl) is not of much use because when we find

its antecedent true than though we are entitled to assert the consequent,

JB*sq, this has no practical significance. It does not matter whose name s

might be, since even the best flesh and blood individual cannot be depended

on not having occasional mental blocks and thus failing to make even the

most obvious inference JBsq does not necessarily follow from JB *sq.

However (dl) is logically equivalent to

(d2) (p – q) & -JBB*sq -+ -Ksp^{4}

The last expression is useful and I propose to provide an important illustration

of its application.

II

Mark Steiner in discussing Descartes’ dream argument points out that

Descartes’ premises do not seem to entail his conclusion. Steiner claims that

the premises with which Descartes starts out are:

(pi): -K-D (I do not know that I am not dreaming)

and

(P2): D -+ -KS (If I am dreaming I do not know that I am

standing up)

and his conclusion

(g): -KS

Now since Descartes has not asserted that D it is not clear how (g) may be

derived from (Pi) and (p2). Steiner claims that (g) may be derived from the

given premises provided we add a third premise not mentioned by Descartes,

namely:

(*) If one is committed to -KP (P for any sentence) then it is irrational

for him to assert P.

His demonstration then proceeds as follows:

(1) KS – -D from (p2) by counterposition

(2) K(KS -+ -D) (1) Necessitation rule of epistemic

logic

(3) KKS -+ K’D (2) Another rule of epistemic logic

(4) -K-D -* -KKS (3) Counterposition

(5) -~-KKS (Pl) and (4) modus ponens

(6) Asserting KS is irrational (5) and (*)

This is ingenious. In fact too much so, thus it is hard to believe that Descartes

‘thought’ of all this in the vaguest sense of that term. I believe that many will

have a strong feeling that there must be some other, shorter way, by which to

arrive at (6). Let me supply an argument showing why this feeling is justified.

We shall suppose that there are two surveyors aand b where a is a highly

intelligent person who among other things is an accomplished logician while b

is a very simple minded person whose numerous limitations include a total

ignorance of formalogic.

CASE 1: Let Da = The calculator used by a is defective

S = The amount of material required to cover the walls

of building B is n.

Let us also suppose that a performs a series of measurements and computa

tions some of which he checks with his calculator and at the end of which he

arrives at the conclusion that S – which happens to be true. It so happens

that a does not know (while b does know) that the computer he is using was

not defective. It is clear then that we have

(pl) -Ka -Da and (p2) Da -*-KaS.

These being parallel statements to those dealt with by Steiner, we need go

into any details in order to show that a will arrive via the steps taken by

Steiner at (5′) -KaKaS.

CASE2: Let us have the previous situation repeated exactly while substituting

b for a and vice versa. In particular b in now using a calculator which

he does not know to be in good order, while a knows this (and soon the role

of this bit of assumption will become clear). Obviously we shall get as far as

(1 “)KbS – -Db, but no further. Steiner has explained that he moved from

(1) to (2) = K(l), on the assumption that if I myself have demonstrated the

truth of (1) then surely not only must (1) be true and I believe it to be true,

but I am justified in believing it to be true, that is, I know that (1) is true i.e.

K(1). Thus the move from (1) to (2) is facilitated not by pure logic alone,

it also requires that I myself should have derived (1) from true premises, that

is from (P2). It follows therefore that for any person who is not aware that

(P2) entails (1) because, for instance, he does not know the rule of logic

called Counterposition, and b is such kind of a person, K(l) is not true and

hence (6) remains underivable. In other words it is impossible to demonstrate

that -KbKbS.

Some may find this result paradoxical. We have seen that a who is a superior

surveyor does definitely not know that S, while b, in spite of the fact

that he is much less likely to do a good job, is in the advantageous position

in which he cannot legitimately be labelled as being ignorant of his knowledge

that S. There will be however philosophers who will not fmd this result intolerable.

In other contexts it has already been remarked that sometimes

“One may know less by knowing more”6 and it is not ultimately so strange

that a sophisticated person should be shrewd enough to discern certain

obstacles that stand in the way of knowing a proposition which may happen

to be true, obstacles to which a more simple minded person remains oblivious.

Hence the former may gain the insight to conclude that he is not entitled

rationally to subscribe to S even though S may well be true while another

person in his blissful ignorance cannot be disqualified from holding that he

knows that S. It was after all Socrates who said that he knew more than

others since he was aware of his total ignorance. It may well be that Socrates

happened to hold a lot of true beliefs,beliefs also held by his fellow Athenians.

Yet he, but not they, disowned the claim that he actually knew that those

beliefs represented the truth.

However from our previous story we may also derive another strange

looking result which cannot so easily be smoothed out. While admittedly

Kb (Kbs — -Db) does not follow from (1 “) the expression (2*) Ka(KbSb -*

-Db) is derivable since a knows enough logic to arrive at (1”). From this,

following Steiner we get

(3 *) KaKbS – Ka -Db

which is the counterpart of (3) and

(4*) -Ka -Db -* -KaKbS

the counterpart of (4). We recall that in the original reasoning Steiner’s next

step was to apply Modus Ponens to (Pl) and (4) to derive

(5) ,KKS

It is however, clearly impossible to make the parallel step here to derive from

(pr) and (4*)

(5*) -KaKbS

since (pr) is -Kb -Db and the antecedent of (4*) is quite a different proposition,

namely, -Ka -Db. Not only are we not given this last proposition but

in fact it was explicitly stipulated that on the contrary Ka -Db for we said

that a knew that the calculator used by b was in working order.

We have thus reached a paradoxical result. We have seen that in Case 1 a

definitely does not know that KaS but in Case 2, which is a mirror image of

Case 1 with the roles of a and b precisely reversed, a himself in spite of all

his knowledge and intelligence is not in the position to declare b to be ignorant

of KbS or that b is ignorant of KbS. The previous explanation does not seem

to be applicable here. After all Socrates who was enlightened enough to hold

that he knew nothing, did not regard his intellectually less accomplished

fellow citizens to be more knowledgeable. On the contrary he was prepared

to declare their ignorance with even greater conviction than his own.

Thus we have added reason to believe that here exists a different deriva

tion of Descartes’ conclusion, in particular, one which is more straight forward

and does not require the troublesome step from (1) to (2), which is not

legitimate, unless the knower himself has established (1). We expect there to

be a briefer route leading to the conclusion one which will also permit the

derivation of -KaKbS no less than -KaKaS.

III

In my book Metaphysics: Methods and Problems ^{7} I have demonstrated that by

applying elementary probability notions, Descartes’ conclusion may easily

be derived from (Pi) and (P2). It seems to me now however that Steiner

would be entitled to reject this derivation. He may very well claim that it is

a misunderstanding of the concept of probability to think that it may be

applied in the context of Descartes’ argument. We are certainly not faced

with a situation concerning which we could say that in the past in similar

instances the frequency of cases that turned out to be such that our experiences

have proved to have been generated by genuine external factors was

such and such. It makes therefore no sense to assign any probabilities to

metaphysical statements in general and to the statement ‘the external world

exists’ in particular.

However, now we are in a position to overcome all difficulties and objections

by making use of the epistemic principle we have introduced in the first

section, namely (d2)(P — q) & -JB*sq -* -Ksp. Descartes asserted (Pi)

-K -D. The reason he gives is that all his experiences are insufficient to

imply, indicate or provide rational basis for believing with enough confidence

that what goes on in his mind is generated by corresponding external

factors any more than this is true in the case of a dream. In other words he

holds (p*’) JB* -D. Now substitute into (d2) ‘KS’ for ‘p’ and ‘-D’ for q

and we get

(d*) (KS ‘ -D) & -JB * -D ‘ -KKS.

Conjoining (1) and (p*’) we get the antecedent of (d*) and that together with

(d2) logically implies by modus ponens -KKS.

With the aid of our theorem we thus are able to derive quickly the desired

conclusion without having to make use of concepts like probability whose

applicability in the present context may be questionable. The brevity of our

proof in comparison to Steiner’s makes it into a somewhat more likely can

didate to amount to reconstruction of Descartes’ own reasoning. An additional

important advantage it has of course that it does not involve Steiner’s

problematic move from (1) to (2) and our proof is therefore independent

of the logical acumen of the knower.

NOTES

1 Notre Dame Journal of Formal Logic (1972), p. 435.

2 Ibid.

3 I owe this important point to the Editor.

4 Another valid expression is

(e) JB*s [p & (p – q)] JB*sq

p may of course be false, but should it be the case that JB*sq then surely there is no

justification to believe both p and p – q which entail q. Obviously also

(f) (p – q) &JBsp & -IEsq – JB*sq

is valid where ‘IEsq’ stands for ‘s has some independent evidence relevant to the credibility

of q’.

5 ‘Cartesian scepticism and epistemic logic’, Analysis (1979).

6 e.g., Carl Ginet, ‘Knowing less by knowing more’, MW Studies in Philosophy (1980),

pp. 151-161.

7 (Oxford, 1983), p. 255.

Department of Philosophy,

University of North Carolina,

Chapel Hill, NC 2 7514,

U.S.A.