The (almost really) Complete Works of Lewis Carroll

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The Method of Trees

Chapter I. Introductory

The essential character of an ordinary Sorites-Problem may be described as follows. Our Data are certain Nullities, involving Attributes, some of which occur both in the positive and in the negative form, and are the Eliminands; while others occur in one form only, and are the Retinoids. And our Quaesitum is to annul the aggregate of the Retinends (i. e. to prove it to be a Nullity).

Hitherto we have done this by a direct Process: that is, we have begun with two of the given Nullities, containing a pair of Eliminands differing only in sign (e. g. a and $a^{\prime }$), and we have treated them as the Premisses of a Syllogism in Fig. I, and have combined them so as to form a new Nullity, not containing the Eliminands: This Partial Conclusion we have then combined, in the same way, with some other given Nullity: and in this way we have proceeded, gradually turning out the Eliminands, till finally we have proved, as our Complete Conclusion, a Nullity consisting of the aggregate of the Retinends.

In the Method of Trees this process is reversed. Its essential feature is that it involves a Reductio ad Absurdum. That is, we begin by assuming, argumenti gratia, that the aggregate of the Retinends (which we wish to prove to be a Nullity) is an Entity: from this assumption we deduce a certain result: this result we show to be absurd: and hence we infer that our original assumption was false, i. e. that the aggregate of the Retinends is a Nullity.

Chapter II. Sorites-Problems with Biliteral Premisses

As the simplest possible example of this Method, let us take the original typical Syllogism in Fig. I, viz. $$x{m}_0†y{m}_0^{\prime }⁋x{y}_0$$

Here our Data are the two Nullities, $x{m}_0$ and $y{m}_0^{\prime }$ involving the Attribute m both in the positive and in the negative form: and our Quaesitum is the Nullity $x{y}_0$.

We begin by assuming that the aggregate $xy$ is an Entity: i. e. we assume that some existing Thing has both the Attributes x and y.

Now the first Premiss tells us that x is incompatible with m. Hence the “Thing” under consideration, which is assumed to have the Attribute x, cannot have the Attribute m. But it is bound to have one of the two m or $m^{\prime }$, since these constitute an Exhaustive Division of the whole Universe. Hence it must have the Attribute $m^{\prime }$.

Similarly, from the second Premiss, we can prove, as our second result, that the “Thing” under consideration has the Attribute m.

These two results, taken together, give us the startling assertion that this “Thing” has both the Attributes, m and $m^{\prime }$, at once; i. e. we get $$x{y}_1⁋xy{m}^{\prime }{m}_1$$

Now we know that m and $m^{\prime }$ are Contradictories: hence this result is evidently absurd: so we go back to our original assumption (that the aggregate $xy$ was an Entity), and we say “hence $xy$ cannot be an Entity: that is, it is a Nullity.”

Now let us arrange this argument in the form of a Tree.

I must explain, to begin with, that all the Trees, in this system, grow head-downwards: the Root is at the top, and the Branches are below. If it be objected that the name “Tree” is a misnomer, my answer is that I am only following the example of all writers on Genealogy. A Genealogical “Tree” always grows downwards: then why may not a Logical “Tree” do likewise?

Well, then, I put the Root of my Tree at the top. It consists of the aggregate $xy$: and the mere writing down of these two Letters is to be understood to mean (using the regular form of a Reductio ad Absurdum) “The aggregate $xy$ shall be a Nullity: for, if not, let it be an Entity; that is, let a certain existing Thing have the two Attributes, x and y.”

Underneath this $xy$ I then place the Letter $m^{\prime }$ (this is part of the Stem of our Tree): and on its left-hand side I place the Number 1, followed by a full-stop, so that our Tree is now

The meaning of this is, that the “Thing,” which is assumed to have the two Attributes x and y, must also have the Attribute $m^{\prime }$: and the Number 1 refers you to the first Premiss as my authority for this assertion.

Next, I place the Letter m on the right-hand side of $m^{\prime }$, and the Number 2, followed by a comma, on the left-hand side of the 1, so that our Tree now is

This means that the Thing must also have the Attribute m (i. e. that $xy{m}^{\prime }m$ is an Entity), and that my authority, for asserting this, is the second Premiss. (Observe that the two Letters, in the lower line, are to be read from left to right, but the two Reference-Numbers from right to left.)

Now we know that $m^{\prime }$ and m are Contradictories: hence it is impossible for an Aggregate, which contains them both, to be an Entity, hence it is a Nullity. And this fact I indicate by drawing a little circle (representing a nought) underneath, so that our Tree now is

The meaning of the circle is “The aggregate of Attributes, beginning at the Root, down to this point, is a Nullity.“

Next, I place, underneath the little circle, the Conclusion $\therefore x{y}_0$, so that the Tree now is

The meaning of the last line is “We have now proved, from the assumption that $xy$ was an Entity, that this aggregate, $xy{m}^{\prime }m$, must be an Entity. But it is evidently a Nullity. Which is absurd. Hence our assumption was false. Hence we have a right to say “Therefore $xy$ is a Nullity.”

I will now exhibit, in one view, the whole Tree, bit by bit, with the meaning of each bit set against it.

All this magnificent machinery, used to prove one single Syllogism, may perhaps remind the Reader of the proverbial absurdity of using a Nasmyth-hammer to crack a nut: but we shall find, when we get a little further in the subject, and begin to deal with more complex Problems, that our machinery is none too costly for the purpose.

My next example shall be a Sorites-Problem, with five Premisses, but still keeping to that childishly simple kind of Premiss (the only kind, as I pointed out in Part I, p. 184, with which the ordinary Logical textbooks venture to deal), the Biliteral Nullity. I will take, from Book VIII, Chapter III, §3, 8 of Part I, the twenty-third Example, viz. $$\begin{array}{ccccccccc}1& & 2& & 3& & 4& & 5\\ b_1^{\prime }{a}_0& †& d{e}_0^{\prime }& †& h_1{b}_0& †& c{e}_0& †& d_1^{\prime }{a}_0^{\prime }\end{array}$$

Here we can easily see, by inspection, that a, b, d, e, are the four Eliminands, and that c and h are Retinends. (As the Reader already knows, we cannot have more than four Eliminands, with five Premisses, though of course the number of Retinends is unlimited.)

I begin by placing $ch$ at the top of the paper, as the Root. And I then look through the Premisses for the Letter c. I find it in No. 4, which tells me that c and e are incompatible. Hence the Thing which I have assumed to have the Attributes c and h, cannot have the Attribute e. Hence it must have the Attribute $e^{\prime }$. And this I express by placing $e^{\prime }$ underneath with the Reference-Number 4 on the left.

The Tree is now

Next, I look for h among the Premisses. I find it in No. 3, which authorises me to say that $b^{\prime }$ is another Attribute that the Thing must have (since it cannot have b). So I place $b^{\prime }$ in the same line with $e^{\prime }$, and its Reference-Number 3, followed by a comma, away to the left.

The Tree is now

Next, I look for $e^{\prime }$ and $b^{\prime }$ among the Premisses. I find them in Nos. 2 and 1, which authorise me to assert that $d^{\prime }$ and $a^{\prime }$ are also necessary Attributes of the Thing; that is, to assert that the whole aggregate $ch{e}^{\prime }{b}^{\prime }{d}^{\prime }{a}^{\prime }$ is an Entity.

The Tree is now

Next, I look for $d^{\prime }$ and $a^{\prime }$ among the Premisses. I find them together, in No. 5, which asserts that the pair $d^{\prime }{a}^{\prime }$ is a Nullity, and therefore authorises me to assert that the whole aggregate $ch{e}^{\prime }{b}^{\prime }{d}^{\prime }{a}^{\prime }$ is a Nullity.

The tree is now

Hence I may write underneath this, $\therefore c{h}_0$, and the Tree is complete.

I now examine the Premisses, to see whether either c or h is given as existing. I find that, in No. 3, h is so given. So I write the full Conclusion thus: $$\therefore c{h}_0†h_1\text{; i.\hspace{0.17em}e. }{h}_1{c}_0$$

I will now exhibit, in one view, the whole Tree, in the same form as in the previous example.

Here it will be well to pause for a moment in order to point out the beautiful fact that this “Tree” argument may be verified, by converting the Tree into a Sorites. And this may be done by the extremely simple rule of beginning at the lower end, and taking the rows of Reference-numbers upwards instead of downwards, viz. in the order 5, 2, 1, 4, 3. The result will be $$\begin{array}{ccccccccc}5& & 2& & 1& & 4& & 3\\ d^{\prime }{a}^{\prime }& †& d{e}^{\prime }& †& b^{\prime }a& †& ce& †& hb\end{array}$$ which proves $c{h}_0$, as the Reader will see for himself, if he will take the trouble to copy it out, and to underscore the Eliminands.

Chapter III. Sorites-Problems with Triliteral and Multiliteral Premisses

The Sorites-Problems, hitherto discussed, have involved Biliteral Premisses only: the admission of Triliteral, and Multiliteral, Premisses introduces a new feature in the construction of Trees, which needs some preliminary explanation.

Suppose we are in the course of constructing a Tree, and have just proved that the existing “Thing,” which we have assumed to possess the Retinends, must also possess the Attribute a. If, on looking up a in the Register, we find a Premiss containing it along with only one other Eliminand, b, of course we conclude, as in the previous Chapter, that, since the “Thing” cannot have the Attribute b, it must have the Contradictory of b, i. e. $b^{\prime }$. But suppose there is no such Premiss: suppose the only one we can find, containing a, contains two other Eliminands, b and c, what conclusion can we draw from this Nullity? We may say, of course, “Since the Thing cannot have the Pair of Attributes $bc$, it must have the Contradictory to it.” But what is the Contradictory to a Pair of Attributes? The simplest way, I think, of answering this question, is to imagine our Univ. divided, by two successive Dichotomies, for these two Attributes. We know that this will give us the four Classes, $bc$, $b{c}^{\prime }$, $b^{\prime }c$, $b^{\prime }{c}^{\prime }$; and that in one of these four the Thing is bound to be; and that it is barred, by the Nullity we have just found, from being in the first of these four Classes. Hence it must be in some one of the other three, which together constitute the Contradictory to the Class $bc$: i. e. it must have some one of the three Pairs of Attributes, $b{c}^{\prime }$, $b^{\prime }c$, $b^{\prime }{c}^{\prime }$.

Now we might, if we liked, state the result in this way, and proceed to consider what would happen in each of these three events. But it would be a cumbrous process. If we were to treat a Quadriliteral Nullity on the same principle, we should have to allow the Thing the choice of seven different events, each of which we should have to investigate separately; and, with a Quintiliteral Nullity there would be fifteen!

But we may easily group these three Classes under two headings: and the simplest way of doing so is to remember that $bc$ is the only one, of these four Pairs of Attributes, which contains neither $b^{\prime }$ nor $c^{\prime }$: i. e., every other Pair contains either $b^{\prime }$ or $c^{\prime }$. Hence we are authorised to say the Thing must have either $b^{\prime }$ or $c^{\prime }$. In other words we may say the Thing must have either the Contradictory of b or the Contradictory of c.

The Reader will easily see that the three possible Pairs, $b{c}^{\prime }$, $b^{\prime }c$, $b^{\prime }{c}^{\prime }$, can be grouped under these two headings. Under $b^{\prime }$ we can place $b^{\prime }c$ and $b^{\prime }{c}^{\prime }$; and under $c^{\prime }$ we can place $b{c}^{\prime }$ and $b^{\prime }{c}^{\prime }$.

This is, of course, a case of overlapping, or what is called “Cross Division,” since $b^{\prime }{c}^{\prime }$ appears under both headings. Now there is no reason to be so lavish of accommodation for this pampered Class $b^{\prime }{c}^{\prime }$: it ought to be quite content with one appearance. So we may fairly say it shall not appear under the heading $b^{\prime }$: that heading shall contain the Class $b^{\prime }c$ only. This result we can secure by tacking on to $b^{\prime }$ the Letter c; so that the two headings will be $b^{\prime }c$ and $c^{\prime }$. Or we may, if we prefer it, say it shall not appear under the heading $c^{\prime }$: that heading shall contain the Class $b{c}^{\prime }$ only. And this result we can secure by tacking on $c^{\prime }$ the Letter b; so that the two headings will be $b^{\prime }$ and $c^{\prime }b$. It is worthwhile to note that, in each case, we tack on, to one of the single Letters, the Contradictory of the other: this fact should be remembered as a rule.

We have now got a Rule of Procedure, to be observed whenever we are obliged to divide our Tree into two Branches, and, instead of saying the Thing must have this one Attribute, we say it must have one or other of these two Attributes.

I will now take some Sorites-Problems containing “Barred” Premisses. We shall find that the Method of Trees saves us a great deal of the trouble entailed by the earlier process. In that earlier process we were obliged to keep a careful watch on all the Barred Premisses, so as to be sure not to use any such Premiss until all its “Bars” had appeared in that Sorites. In this new Method, the Barred Premisses all take care of themselves: and we shall see, when we come to “verify” our Tree, by translating it into Sorites-form that no Barred Premiss will venture to make its appearance until all its Bars have been duly accounted for.

My first example shall be $$\begin{array}{ccccccccccccccc}1& & 2& & 3& & 4& & 5& & 6& & 7& & 8\\ d^{\prime }{n}_1^{\prime }{m}_0^{\prime }& †& k{a}_1^{\prime }{c}_0^{\prime }& †& l{e}_1{m}_0& †& d{h}_1{k}_0^{\prime }& †& h^{\prime }l{a}_0^{\prime }& †& h{m}_1^{\prime }{b}_0^{\prime }& †& a^{\prime }b{n}_0& †& a{m}_1^{\prime }{e}_0\end{array}$$

Here we see that some of the Letters occur more than once: for instance, h occurs in Nos. 4 and 6, in the positive form, and in No. 5 in the negative form. Hence, when we ask the question, as to any particular Letter, “In which of the Premisses does it occur?”, we should have to interrupt the construction of our Tree, in order to hunt through the whole Set of Premisses. To avoid this necessity, it will be convenient to draw up, once for all, a “Register of Attributes,” from which we get, at a glance, the required information. The rule, for making such a Register, is as follows:

At the left margin of the paper draw a short vertical line, and above it, a little to the right, place the letter a: and under it place two rows of numbers, the upper row referring to the Premisses where a occurs in the positive form, and the lower to those where it occurs in the negative form: then draw another short vertical, to divide the a’s from the b’s, write b over the next space, and proceed as before.

Thus, in the present example, after drawing the first vertical and writing a above, we look through the Premisses, to see which of them contain a or $a^{\prime }$. In No. 2, we find $a^{\prime }$: so we write 2 in the lower line: in Nos. 5 and 7, we find two more: so we write 5, 7, still in the lower line: lastly, in No. 8, we find a: so we write 8 in the upper line: then we draw another vertical, and write b over the next space. The beginning of the Register will now be

I recommend my Reader to copy out these seven Premisses at the top of a large sheet of paper, and underneath them to construct a Register of Attributes for himself, which he can then compare with the one here given, to satisfy himself that he has made no mistake. The Register is as follows:

This result we had better verify, before going further, by the following rule:

Name the Letters in No. 1, in alphabetical order: then look them up in the Register, and see that 1 occurs in its proper place under each. Then name the Letters in No. 2: and so on.

Thus, in this example, we look at No. 1, and say (naming the letters in alphabetical order) “d-dash, n-dash, m-dash.” Then we look up d, n, and m in the Register, and satisfy ourselves that each of them has a 1 under it in the lower line. Then we look at No. 2, and say “a-dash, c-dash, k,” and proceed as before.

This Register not only enables us to see, at a glance, in which Premisses any particular Letter occurs; but it also tells us that this Sorites-Problem contains seven Eliminands (every Letter, that has numbers under it in both rows, is an Eliminand), and three Retinends. It also tells us that there are three Barred Premisses; since, under a, we see that No. 8 is barred by Nos. 2, 5, and 7; under h, that No. 5 is barred by Nos. 4 and 6; and, under m, that No. 3 is barred by Nos. 1, 6, and 8. But these are now trifles, about which we need not trouble ourselves!

In working this Tree, I shall adopt a new plan, which I think the Reader will find beautifully clear and intelligible. Instead of exhibiting the Tree, piecemeal, as I proceed, I shall simply give my soliloquy as I work it out, with the “stage-directions” (given in italics, between square brackets) showing what I $do$: and, if the Reader will simply take a piece of paper, and pen and ink, and will copy, at the top of his paper, the eight Premisses and the Register, and will then, while reading my soliloquy, follow the stage-directions, and thus do all the things himself, he will find that he has constructed the Tree for himself: and he can then, for his own satisfaction, compare his finished result with mine. (Note that the letters [R.R.] will be used to represent the stage-direction I refer to Register.)

My soliloquy is as follows:

“So! Eight Premisses, and every one of them triliteral! However, there are seven Eliminands: so there ca’n’t be any superfluous Premisses. Well, the Conclusion ought to be $c^{\prime }e{l}_0$, of course.”

“Now, what can we do with $c^{\prime }$?”

“It occurs in 2 only: and that tells me that it ca’n’t be $k{a}^{\prime }$: so of course it must be (taking them in alphabetical order) a or $k^{\prime }$. That would force me to divide the Tree at the very Root! Let’s try e.”

“It occurs in 3 and 8: and in 3 it is kind enough to have another Retinend with it, and only one Eliminand! Well, this Premiss tells us that $el$ ca’n’t be m: so of course it must be $m^{\prime }$. Well, there’s one Letter for the Stem, at any rate!”

“Let’s see if l gives us any other certainty for the second row.”

“No! No. 5 is the only other Premiss: and that would ‘divide’ between a and h. We must go on to the third row. What will $m^{\prime }$ do for us?”

“Hm! There’s good choice here! Nos. 1, 6, and 8. No. 1 divides between d and n. No. 6 divides between b and $h^{\prime }$. No. 8 is more gracious: we’ve got both $m^{\prime }$ and e already: so this gives us $a^{\prime }$.”

“Well, now, what will $a^{\prime }$ do for us?”

“Again we have ample choice! No. 2 does beautifully, as we’ve got c upstairs: so that gives $k^{\prime }$ for the fourth row.”

“Any more results from $a^{\prime }$?”

“Yes. No. 5, $h^{\prime }l{a}_0^{\prime }$, and we’ve got l upstairs: so that gives us h.”

“Any more? No. 7 is the other one: and that would have to divide, as we haven’t got either b or n upstairs: so we’ll let it alone. Now for the fifth row. What will $k^{\prime }$ do for us?”

“No. 4 is the only one: and that will do grandly, as we’ve got both h and $k^{\prime }$: so it gives us $d^{\prime }$ as a certainty.”

“And what will h do for us?”

“It occurs in Nos. 4 and 6. But we’ve just used No. 4. Let’s try No. 6. Yes, that gives us b, as we have $m^{\prime }$ upstairs.”

“Now for the sixth row. What will $d^{\prime }$ do?”

“No. 1’s the only one: that gives us n to follow, as we’ve got $m^{\prime }$ upstairs.”

“And will b do us any good?”

“Yes, b gives us $n^{\prime }$, as we’ve got $a^{\prime }$ upstairs.”

“Come! that finishes the thing: $n{n}^{\prime }$ is an absurdity!”

“So now we’ve proved $c^{\prime }e{l}_Q$. The next thing is to examine the Premisses, and see if any of these three are given as existing.”

“$c^{\prime }$ occurs in No. 2 only—non-existent: e occurs in Nos. 3 and 8—and exists in No. 3, along with l. So we get $c^{\prime }e{l}_0†l{e}_1$; that is, $l{e}_1{c}_0^{\prime }$; that is, All $le$ are c; and my task is done!”

Here ends my soliloquy. If the Reader will now turn to the image above, he will see what the Tree ought to look like: and, between that and the Tree he has constructed for himself, I hope he will find a considerable family-likeness! He should then verify his Tree, by writing out the eight Premisses in the reverse order (i. e. in the order 1, 7, 4, 6, 2, 5, 8, 3), omitting all subscripts, and underscoring whatever letters he can eliminate: and the final result ought to be $c^{\prime }e{l}_0$.

My second example shall be $$\begin{array}{ccccccccccccc}1& & 2& & 3& & 4& & 5& & 6& & 7\\ h{m}_1{k}_0& †& d^{\prime }{e}^{\prime }{c}_0^{\prime }& †& h{k}^{\prime }{a}_0^{\prime }& †& b{l}_1{h}_0^{\prime }& †& c{k}_1{m}_0^{\prime }& †& h{c}^{\prime }{e}_0& †& b{a}_1{k}_0^{\prime }\end{array}$$

I will now construct the Tree, soliloquising as I do so.

“Six Eliminands, are there? And seven Premisses—none too many. And three Retinends, b, $d^{\prime }$, and l. Well, those will make the Root.”

“Now, then, what will b do for us?”

“No. 4—why, that gives us a certainty at once! b and l are both of them Retinends.”

“No. 7?”

“Divides. Now for $d^{\prime }$. No. 2?”

“Divides. And now for l. No. 4? We’ve got it already. So that ends our second row. Now for the third. What will h do? No. 1?”

“Divides. No. 3?”

“Ditto. No. 6?”

“Ditto. We must divide, this time: let’s go back to No. 1: ‘first come, first served,’ you know.”

“Now, shall we tack an m on to the $k^{\prime }$? Or shall we tack a k on to the $m^{\prime }$? Let’s see if either of them would be of any future use.”

“Well, m only occurs in No. 1, and that we’ve just used: so $m^{\prime }$ can be of no further use: but k occurs in No. 5 also: so perhaps it may be of use, further down.”

Here I cease to soliloquise, for a moment, in order to inform my Reader that the meaning of this division of the Tree into two Branches is to assert that the (supposed) existing “Thing,” which has the Attributes $b{d}^{\prime }lh$, must also have either the single Attribute $k^{\prime }$ (which it may follow up with m or with $m^{\prime }$, whichever it likes), or else the Pair of Attributes $m^{\prime }k$. I resume my soliloquy.

“Now, in the left-hand branch, what will $k^{\prime }$ do for us?”

“It occurs in No. 3. That’ll do very nicely: we’ve got h and $k^{\prime }$ already, down this Branch: so that gives us a.”

“It also occurs in No. 7: and this gives us another certainty, as we’ve got b upstairs: so this gives us $a^{\prime }$.”

“Well, that Branch is annulled, anyhow!”

“Now for the right-hand Branch. What will $m^{\prime }$ do for us?”

“It occurs in No. 5 only. However, that gives us a certainty, as we’ve got both k and $m^{\prime }$: so we must have $c^{\prime }$ to follow.”

“Now, what will $c^{\prime }$ lead to?”

“It occurs in No. 2, and in No. 6. In No. 2, it gives us e to follow, as we’ve got $d^{\prime }$ upstairs; and, in No. 6, it gives us $e^{\prime }$ to follow, as we’ve got h upstairs.”

“Well, that annuls the right-hand Branch: so the Tree is finished!”

“So now we’ve got $b{d}^{\prime }{l}_0$: let’s see which of them exist in the Premisses.”

“No. 4 gives us $bl$ as existing: that’ll do very well.”

My reader may now refer to the Tree, given above, and see if he has drawn his correctly.

Observe that this Tree, though not containing a single word of English, expresses symbolically the whole of the following argument.

If possible, let $b{d}^{\prime }l$ be an Entity, i. e. let there be a certain existing Thing, which has all three Attributes. Then, by No. 4, this same Thing must also have the Attribute h. Hence, by No. 1, it must also have either $k^{\prime }$ or $m^{\prime }k$. If it chooses $k^{\prime }$, then, by Nos. 3 and 7, it must also have $a{a}^{\prime }$, which is absurd: if it chooses $m^{\prime }k$, then, by Nos. 5, 2, and 6, it must also have $c^{\prime }e{e}^{\prime }$, which is absurd.

More briefly, if an existing Thing has the Attributes $b{d}^{\prime }l$, it must also have either $h{k}^{\prime }aa$ or $h{m}^{\prime }k{c}^{\prime }e{e}^{\prime }$. But each of these aggregates is impossible.

Hence $b{d}^{\prime }l$ cannot be an Entity.

Therefore it is a Nullity.

Here ends my soliloquy; and there is no logical necessity to do anything more: still it is very satisfactory to “verify” the Tree, by translating it into Sorites-form. There will be two Partial Conclusions, which I shall number as 9 and 10. But I must pause here, to instruct my Reader how to deal with Branches, in verifying a Tree. The simple Rule is, when there are two Branches, of which one is headed by a single Letter, and the other by a Pair, to take the single Letter first, turn it into a Sorites, and record its Partial Conclusion: then take the double-Letter Branch: turn it also into a Sorites—but there’s no need to record its result, as we may go on at once with the Premiss used in the Branching: then take the recorded result of the single-Letter Branch: then we can go “upstairs,” if there is any Stem leading down to the Branching. Thus, in the present instance, of the two main Branches, we take $k^{\prime }$ first. So our first Sorites consists of 3 and 7. So we draw a small square against $k^{\prime }$, on the right side of it; and in that square we write 8. Our final Sorites will begin, in the $m^{\prime }k$-Branch, with Nos. 2, 6, and 5. This takes us up to $m^{\prime }k$. Then we cross the bridge, by means of No. 1: then take in No. 8: then we can go upstairs, and take No. 4: and that ought to give us the desired Conclusion.

The following summary exhibits these two Soriteses in a handy form: $$3,7⁋h{k}^{\prime }{b}_0\dots (8);2,6,5,1,8,4⁋d^{\prime }b{l}_0$$

The Reader should satisfy himself that this is correct, by copying the above, substituting, for the reference-numbers, the actual Premisses, and underscoring all the Eliminands. The result ought to be as follows: $$\begin{array}{cccl}3& & 7& \\ h{k}^{\prime }\underset{\_}{a^{\prime }}& †& b\underset{\_}{\underset{\_}{a}}{k}^{\prime }& ⁋h{k}^{\prime }{b}_0\dots (8);\end{array}$$

$$\begin{array}{cccccccccccl}2& & 6& & 5& & 1& & 8& & 4& \\ d^{\prime }\underset{\_}{e^{\prime }}\underset{\_}{c^{\prime }}& †& \underset{\_}{h}\underset{\_}{c^{\prime }}\underset{\_}{\underset{\_}{e}}& †& \underset{\_}{\underset{\_}{c}}\underset{\_}{k}\underset{\_}{m^{\prime }}& †& \underset{\_}{h}\underset{\_}{\underset{\_}{m}}\underset{\_}{k}& †& \underset{\_}{h}\underset{\_}{\underset{\_}{k^{\prime }}}b& †& bl\underset{\_}{\underset{\_}{h^{\prime }}}& ⁋d^{\prime }b{l}_0\end{array}$$

I will now work out a rather harder Problem.

Let us take the Sorites, $$\begin{array}{ccccccccccccccc}1& & 2& & 3& & 4& & 5& & 6& & 7& & 8\\ kn{l}_0^{\prime }& †& c{h}_1{e}_0^{\prime }& †& b{l}^{\prime }{a}_0^{\prime }& †& d_1{e}^{\prime }{n}_0^{\prime }& †& ah{c}_0^{\prime }& †& n{b}_1{k}_0^{\prime }& †& l{e}_1^{\prime }{m}_0& †& d^{\prime }{h}^{\prime }{n}_0^{\prime }\end{array}$$

There are eight Premisses, seven Eliminands, and three Retinends.

My soliloquy is as follows:

“No superfluous Premisses, this time. Two Barred Premisses, and a Barred Group! But no matter: the Tree will take care of all that!”

“Now, what will b do for us?”

“In 3 it divides: and in 7 it divides. Let’s try $e^{\prime }$.”

“In 2 it divides; but 7 suits us better: both $e^{\prime }$ and m are Retinends: so the Thing, that’s got all the Retinends, ca’n’t be l, and therefore must be $l^{\prime }$.”

“Will m give us any more certainties?”

“No: it only occurs in 7, which we’ve just used. Now, what will $l^{\prime }$ do for us?”

“It occurs in 1 and 3. In 1, it divides: but we’ve better luck in 3, as b’s a Retinend. So that gives us a to go on with.”

“Now for a.”

“It occurs in 5 only: and that divides. Well, there’s no help for it, this time! We must divide between c and $h^{\prime }$.”

“Now, we’ve got the right to tack on $c^{\prime }$ to the $h^{\prime }$, or h to the c, whichever we like. Shall we do either? Let’s see if either of them would be of any use, further down.”

“$c^{\prime }$ is no use: it only occurs in 5, the one we’re using. But h occurs also in 2: so we’d better tack it on.”

“Now what will c, or h, do for us?”

“c occurs in 2—and h along with it: and $e^{\prime }$ is a Retinend: so that gives us a Nullity at once.”

“Now, what can we do with $h^{\prime }$?”

“Well it divides in 8; and I’m afraid there’s no help for it, as that’s the only one it occurs in.”

“Now shall we tack on $d^{\prime }$ to n? Or $n^{\prime }$ to d? Let’s see if $d^{\prime }$ could be of any use further down.”

“No, it couldn’t. Could $n^{\prime }$?”

“Yes, it might. Very well, then we’ll tack it on, on the chance.”

“Well, there’s no use going back to the left-hand Branch: it’s extinct. So we must go on with this one. Will d help us at all?”

“Yes! It occurs in 4, along with $n^{\prime }$ and a Retinend. So here we get another Nullity!”

“Now there’s only one Branch left to attend to. What can we do with n?”

“n occurs in 1 and 6. In 1 it gives us $k^{\prime }$, as we’ve got $l^{\prime }$ upstairs: and in 6 it gives us k, as we’ve got b upstairs. And $k^{\prime }k$ is an obvious absurdity. So this brings the whole thing to an end.”

“Well, that proves $b{e}^{\prime }m$ to be a Nullity. But do any of them exist separately?”

“Yes, each one exists, by itself: but we’re not told that any two exist together. Well, let’s make b exist, then.”

“So now the Tree is in full leaf!”

The Reader can now look above, and compare his Tree with the one there depicted.

The Verification of this tree shall now be given in a second soliloquy:

“Well, now to verify our result. Where are the Partial Conclusions to come? At the first Branching, of course we must take $h^{\prime }$ first: and, at the second Branching, we must take n first. So the first Partial Conclusion must be at n: and it must be No. 9, as we’ve got eight Premisses.”

“That first Sorites consists of Nos. 1 and 6. Then, for the second Sorites, we must take the two-Letter Branch—the $d{n}^{\prime }$-Branch. So we take No. 4: then cross the bridge with 8: then take in 9: then we go upstairs, and record the result as No. 10.”

“Then, for the final Sorites, we must begin with 2: then cross the bridge with 5: then take in 10; then we go upstairs, and take 3 and 7: and that ought to prove $b{e}^{\prime }{m}_0$.”

The Reader should now write out these three Soriteses, in full according to the following summary, and do all the necessary underscoring, and thus satisfy himself that they really do prove the Conclusion.

$1,6⁋n{l}^{\prime }{b}_0\dots (9)$; $4,8,9⁋e^{\prime }{h}^{\prime }{l}^{\prime }{b}_0\dots (10)$; $2,5,10,3,7⁋e^{\prime }b{m}_0$

I will now take a still harder Problem, and solve it in the same way.

$$\begin{array}{cccccccccccccc}& 1& & 2& & 3& & 4& & 5& & 6& & 7\\ & a{n}_1{b}_0^{\prime }& †& w{m}_1{l}_0^{\prime }& †& cs{n}_0& †& a{r}_1{v}_0^{\prime }& †& e_1{c}^{\prime }{l}_0^{\prime }& †& m{h}_1{t}_0^{\prime }& †& k_1{n}_0^{\prime }\\ & 8& & 9& & 10& & 11& & 12& & 13& & 14\\ †& d{r}_1{a}^{\prime }{e}_0^{\prime }& †& r{l}_1{w}_0^{\prime }& †& e{l}_1^{\prime }{n}_0^{\prime }& †& a^{\prime }{s}_0^{\prime }& †& d{b}_1{m}_0^{\prime }& †& v_1{e}^{\prime }{k}_0^{\prime }& †& b{w}_1^{\prime }{h}_0^{\prime }\end{array}$$

There are fourteen Premisses, twelve Eliminands, and three Retinends.

The Reader should now take a large sheet of paper, and copy the above fourteen Premisses at the top: then put the book aside, and make and verify his own Register: then compare it with mine: then copy the words, “There are fourteen &c.”: and then he will be able to understand the following soliloquy:

“Fourteen Premisses, and only twelve Eliminands? There may be a superfluous Premiss. And three Retinends.”

“Now for d. No. 8? It occurs there, along with another Retinend, l; but, even with that help, it has to divide. Let’s try the other Premisses containing Retinends. They are Nos. 12, 2, 5, 10, 4, 8, and 9. No, it’s no use! They all divide! So let’s go back to No. 8.”

“Now, we may tack on $a^{\prime }$ to e, or $e^{\prime }$ to a, whichever we like. Will either of them do any good? Well, $a^{\prime }$ might be used further down—and so might $e^{\prime }$. Then it doesn’t matter which we take. Let’s move from left to right—moving the other way would seem like writing backwards!”

“Now, what does a give us? It occurs in 1 and 4. In 1, it divides. But, in 4, it gives us v, as we’ve got r upstairs.”

“Now for the right-hand Branch. Is e of any use? It’s in 5, and 10. In 5, it gives us c, in 10, it gives us n.”

“And will its partner, $a^{\prime }$, help us? It occurs in 8 and 11; but of course 8 is no good, as we’ve used it in the Branching. However, 11 gives us another Letter s: so we’ve actually landed three fish in one haul this time!”

“Now we go back to v. Well, that only occurs in 13: so it’s got to divide, I’m afraid!”

“Now, is it worthwhile tacking on an $e^{\prime }$ or a $k^{\prime }$? I see $e^{\prime }$ occurs in 8; but we couldn’t use 8, down this Branch, as it would want $a^{\prime }$, and we’ve got an a upstairs: so that’s no good. Where does $k^{\prime }$ occur? Nowhere else, besides 13, I see. Then there’s no use tacking on either. So we’ll let them alone. Now we go back to our grand haul, $cns$. Where does c occur? No. 3? Why, that actually slays all three at once!”

“Now we return to e. Let’s see: we’ve had e somewhere before. Oh, there it is, in the right-hand Branch! So this e can perhaps make use of the annulment of the earlier one, provided that the other e didn’t need its partner, $a^{\prime }$, to help to annul it, since this e has got a as a partner. Did it need it? What Premisses does $a^{\prime }$ occur in? Nos. 8 and 11. And was either of them used in the annulment? Yes, we used 11. Then I’m afraid this new e ca’n’t get any help from the old one. It must manage its own annulment. What can we do with it? It occurs in 5 and 10. In 5, it gives us c: in 10, it gives us n.”

“But we ca’n’t tack on an s, this time, as we haven’t got an $a^{\prime }$ to help us! Let’s go to the k-Branch. What will k do? It occurs in 7 only: and that gives us n.”

“Now back to the left-hand again. What will c do? It occurs in 3—along with n luckily: so that gives us $s^{\prime }$.”

“And will c’s partner, n, do anything for us? Yes in 1, it gives us b, as we’ve got an a upstairs.”

“Now back to the k-Branch. What can we do with n? Why, we’ve got another n, on the same level, in the e-Branch! So this one had better wait on the chance of being able to avail itself of the annulment of the other.”

“Now to the left again. What can we do with $s^{\prime }b$? Well, $s^{\prime }$ only occurs in 11, and that needs an $a^{\prime }$: so $s^{\prime }$ gives us no assistance. Will b do any good? It occurs in 12 and 14. In 12, it gives us m: in 14, it divides.”

“Any other Branch to go to? No, the other one is waiting: we must stick to this one till it’s finished. What can we do with m? Well, it occurs in 2 and 6. In 2, it gives us $w^{\prime }$: in 6, it divides.”

“Now for $w^{\prime }$. It occurs in 9 and 14. In 9, it gives us $t^{\prime }$: in 14, it gives us h.”

“Now, what will $t^{\prime }$ do? It occurs in 6 only: but there it comes along with h, and we’ve got m upstairs: so that annuls this Branch.”

“Now, we’ve got an n in the other Branch, patiently waiting to learn the fate of its namesake on this Branch. So, now that this n has got itself annulled, the question is whether the waiting n can use the same annulment, in which case we need only refer to it, without taking the trouble to write it out again. Now this new n has the same ancestors as the old n, with the exception of its brother c, and its father e. So, if the left-hand n managed to get annulled without using either of these two kinsmen, then its annulment will serve for the right-hand one: if not I’m afraid the new n must devise an annulment of its own. Now, was c or e used in that annulment?”

“Yes! c was used in the very next row! it gave us $s^{\prime }n$. So this n will have to devise an annulment for itself—no, stay! That $s^{\prime }$ was of no further use! So, after all, c was not used in the annulment. Well, then, was e used?”

“No, e only occurs in 5 and 10; and neither of those was used in the annulment. So, after all this new n can use the old annulment.”

“So now the Tree is complete, and we’ve proved the Nullity $d{l}^{\prime }{r}_0$. Now, are any of these Letters given as existent? Let’s see.”

“Yes, $dr$ exist, together, in No. 8.”

Here ends my soliloquy. But we had better verify our result, by translating our Tree into Sorites-form. This shall be done in a supplementary soliloquy.

“Well, now to verify this Tree. And first, what reference-number must be put under the n that we kept waiting so long? To answer this question, we must first settle in what order we’re going to take the Soriteses that are to prove our Partial Conclusions. Let’s see. At the first Branching, of course it’s the a-Branch that must be proved first, as it’s a single-Letter one, and the other is a double-Letter one. At the second branching, both are single-Letters: but of course we must take the e-Branch first, as it’s the only one we can prove, to begin with, since it ends in a circle. So the first Sorites must run up as far as $cn$, and then record its result, for the benefit of the waiting n: that Sorites will consist of Nos. 6, 9, 14, 2, 12, 3, and 1. But wait a moment! Will it contain No. 3? No, of course it wo’n’t! No. 3 only served to give us $s^{\prime }$, and $s^{\prime }$ turned out to be useless! Then the first Sorites will simply be 6, 9, 14, 2, 12, and 1. And we must call its result No. 15, as there are fourteen Premisses.”

“Then the second Sorites had better take in the whole of the e-Branch, and record its result at the top. So it will be a very short one—merely containing 15 and 10: of course missing 5, as that was only wanted for the useless c.”

“Then the third Sorites will have to work its way up the k-Branch. That is, it must begin with 15: then take 7: then cross the bridge, by means of 13: then take in 16: then go upstairs and take 4: and then we shall have to record its result.”

“And the final Sorites must of course run up the $e{a}^{\prime }$-Branch. So it will begin with 3, 5, 10, 11: then cross the 8 bridge: then take in 17: and that ought to finish the thing, as there’s no stem above that first Branching. So the four Soriteses will run as follows:

$6,9,14,2,12,1⁋15$; $15,10⁋16$; $15,7,13,16,4⁋17$; $3,5,10,11,8,17⁋d{l}^{\prime }{r}_0$.”

The Reader should now write out these Soriteses, in full, and do all the necessary underscoring, and satisfy himself that they do really prove the desired Conclusion.

I will now go through a really long and hard Problem of this kind, soliloquy fashion, and I think that the Reader, if he has the patience to work it through, taking my soliloquy as his guide, will then find himself fully competent to solve any ordinary Sorites-Problem: those, that have special features, will be considered in subsequent chapters.

The twenty-four Premisses of this Problem are as follows: $$\begin{array}{cccccccccc}1& & 2& & 3& & 4& & 5& \\ C{l}_1{E}_0^{\prime }& †& A{v}_1^{\prime }{D}_0& †& k_1{m}_0^{\prime }& †& l{C}_1^{\prime }(b^{\prime }{n}^{\prime }{)}_0^{\prime }& †& ds{b}_1{t}_0^{\prime }& †\\ 6& & 7& & 8& & 9& & 10& \\ t{D}_1{w}_0^{\prime }& †& d{r}^{\prime }{a}_1^{\prime }{A}_0^{\prime }& †& v{w}_1{B}_0& †& e{m}_1^{\prime }(r^{\prime }{b}^{\prime }{)}_0^{\prime }& †& H{a}_1{c}_0^{\prime }& †\\ 11& & 12& & 13& & 14& & 15& \\ dtma{v}_0^{\prime }& †& ds{t}_1{A}_0^{\prime }& †& D{n}^{\prime }{r}^{\prime }{b}_1^{\prime }{z}_0& †& c{E}^{\prime }{z}_0& †& b{s}_1^{\prime }{l}^{\prime }{e}_0^{\prime }& †\\ 16& & 17& & 18& & 19& & 20& \\ at{E}_1{v}_0^{\prime }& †& rD{h}_1^{\prime }{e}_0^{\prime }& †& m{t}_1^{\prime }{D}_0& †& An{l}_1^{\prime }{c}_0^{\prime }& †& rd{k}_1^{\prime }{h}_0& †\\ 21& & 22& & 23& & 24& & & \\ zt{B}_1^{\prime }{d}_0& †& n{l}_1^{\prime }{H}_0^{\prime }& †& E{t}_1^{\prime }{z}_0& †& dzr{A}_1^{\prime }{a}_0^{\prime }\end{array}$$

Before making the Register, it may be well to point out that No. 4 means “All $l{C}^{\prime }$ are $b^{\prime }{n}^{\prime }$”; i. e. “All $l{C}^{\prime }$ are $b^{\prime }$, and all $l{C}^{\prime }$ are $n^{\prime }$.” Hence this Premiss really contains two distinct Propositions, which we might, if we chose, symbolise as $l{C}_1^{\prime }{b}_0†l{C}_1^{\prime }{n}_0$ (so that b and n must be reckoned as appearing in the positive form in this Premiss). If I have to use the whole Premiss at once, I shall refer to it as 4, simply; but, if I have to use either part by itself, I shall refer to it as 4*, or as 4**. Similar remarks will apply to No. 9. Hence the actual number of Premisses is twenty-six.

I recommend the Reader to copy these Premisses at the top of a large sheet of paper, and then to make the Register for himself, without looking at mine; then to verify it, by the method he has already learned (see p. 285); and lastly to compare it with the Register here given.

My soliloquy is as follows:

“Twenty-six Premisses, nineteen Eliminands, and three Retinends, d, z, and D. So there are six extra Premisses. Looks as if there might be some superfluous ones: and perhaps a Retinend might be spared: let’s try.”

“No: there seems no chance of getting rid of a Retinend. So now for our Tree.”

“Now what can we do with d? It occurs in 5, 7, 11, 12, 20, 21, 24. Alas, they all divide! And so do the z’s: and so do the great D’s. Well, there’s no help for it: we must divide at the very first start! Let’s get a biliteral division, if we can. No. 21 is the first I can find, as it contains two Retinends: so it merely divides for t and $B^{\prime }$.”

“Now, is there any use tacking on t or $B^{\prime }$? Let’s see. Yes, t can be of further use, but $B^{\prime }$ of none.”

“Now, for the $t^{\prime }$-Branch. 5 divides, but 18 doesn’t: it gives us $m^{\prime }$. And 23 gives us $E^{\prime }$. That’s a good beginning.”

“Now for the $Bt$-Branch. B only occurs in 8; and that divides. However, t helps us in 6, and gives us w: in all the other Premisses it divides.”

“Now we go back to the $t^{\prime }$-Branch. What will $m^{\prime }$ and $E^{\prime }$ do for us? $m^{\prime }$ occurs in 3, and that gives us $k^{\prime }$. In 9 it divides, even if we take 9 piecemeal. $E^{\prime }$ divides in 1, but in 14 it gives us $c^{\prime }$. That’ll do capitally.”

“Now for the $Bt$-Branch again. What will w do? It occurs in 8, and gives us $v^{\prime }$.”

“Now we go back to the $t^{\prime }$-Branch again. What can we do with $k^{\prime }$ and $c^{\prime }$? $k^{\prime }$ only occurs in 20, and that divides, $c^{\prime }$ occurs in 10 and 19, but they both divide. Then we will take $k^{\prime }$: that will give us $h^{\prime }$ and $r^{\prime }$ for our Branches.”

“Now would either h or r be of any further use? h wo’n’t, but r occurs in three other Premisses.”

“Now back to the $Bt$-Branch. What will $v^{\prime }$ do? It occurs in 2, 11, and 16. In 2 it gives us $A^{\prime }$. In 11 and 16 it divides.”

“Now back to that last Branching. What will $h^{\prime }r$ do? $h^{\prime }$ occurs in 17; and that gives us e at once, as we’ve got three of the four letters already. And r occurs in 9 (which we must break up, and take $e{m}^{\prime }{r}_0$ by itself), and that gives us $e^{\prime }$. No use troubling about 24: we’ve got our Nullity already.”

“Gome, there’s one Branch annulled already! The $r^{\prime }$-Branch is the only one we have to go on with, at present. Let’s see what $r^{\prime }$ does for us. It occurs in 7 and 13, and both divide. Let’s take 7.”

“Now, would $a^{\prime }$ or $A^{\prime }$ be of further use? Well, $a^{\prime }$ occurs in 24; but there it wants $A^{\prime }$ as a partner, which of course it can’t have: so it’s no use. Great $A^{\prime }$ occurs in 12 and 24; but in 12 it wants t, which it can’t have; and 24 we know to be useless. So there’s no tacking on to be done, this time! Now we go back to $A^{\prime }$. In 7 it divides: in 12 it gives us $s^{\prime }$: in 24 it divides.”

“Now back to the left again. What will a do for us? In 10 it gives us $H^{\prime }$, as we’ve got $c^{\prime }$ upstairs. We can’t use 11, as it wants t, and we’ve got $t^{\prime }$ upstairs: and 16 wants E, and we’ve got $E^{\prime }$ upstairs: so 10’s the only one.”

“Now for the A-Branch. A gives us v in 2: in 19 it divides: $a^{\prime }$ occurs only in 24 (besides 7, which made the Branching) and there it wants $A^{\prime }$: so we can’t use it.”

“Now away to the right again. What will $s^{\prime }$ do?”

“It occurs only in 15, and there, alas, it divides into three Branches! That’s a very cumbrous process, and a thing to be avoided as long as possible. So let’s draw a double-line under $s^{\prime }$, to show that we’ve rejected its guidance for the present, and ‘hark back’ for something that will divide into two Branches.”

“Now, will $A^{\prime }$ serve our purpose? Yes, that’ll do very well: in 7 it divides into a and r. And we must remember, in case we succeed in annulling this Branch, to examine whether we’ve used this $s^{\prime }$ anywhere below; for, if not, No. 12 will be a superfluous Premiss—unless it happens to be used in the left-hand Branch.”

“Now, would $a^{\prime }$, or $r^{\prime }$, be of any further use? Yes, $a^{\prime }$ could be used in 24: that will do.”

“And $r^{\prime }$ could be used in 13. Which will be best? I see that a has appeared before. Now we go back to the left. What will $H^{\prime }$ do? It occurs only in 22; and there it divides. This is a very branchy Tree!”

“Now, will $l^{\prime }$ or n be of further use? Yes, each of them might. $l^{\prime }$ occurs in 15 and 19; and neither of those demand impossible partners. And n occurs in 4 and 19. In 4 we could use it, as it wants l for a partner; but not in 19, as there it demands $l^{\prime }$. Well, it’s arbitrary which we tack on: let’s keep l as the single Letter.”

“Now for the other Branch. What can we do with v? Well, it occurs only in 8. So we’ve no choice.”

“Would w or B be of further use? No, neither of them. So we go away to the right again, and try our luck with the $Bt$-Branch. What can we do with a? It occurs in 10, 11, and 16. In 10, it divides: but in 11 it gives us $m^{\prime }$: and in 16 it gives us $E^{\prime }$.”

“Now for $r{a}^{\prime }$. What will r do? In 9 it divides: in 17, ditto: in 20, ditto: but in 24 it gives us a Nullity!”

“Now we go back to the extreme left-hand again, and take the first Branch we find, that’s still growing. What will l do for us? In 1, it gives us $C^{\prime }$. No. 4 we ca’n’t use, yet; though we shall be able to, next time we come this way.”

“Now for $n^{\prime }{l}^{\prime }$. $n^{\prime }$ occurs in 13, which looks alarming, it’s so full of Letters: however, we’ve got all but one, upstairs! So that gives us b: $l^{\prime }$ occurs in 15; but that would divide. It also occurs in 19; but there it wants n for a partner. Well, we’ve got one Letter, anyhow!”

“Now for $w^{\prime }$. Well, $w^{\prime }$ occurs only in 6: and there it wants t for a partner, and ca’n’t have it! So this Branch wo’n’t grow any further. Will the $B^{\prime }$-Branch be more vigorous? No, not a bit of it! It only occurs in 21, and there it demands t for a partner! So both these Branches come to a deadlock. Well, there’s nothing for it but to draw a double-line under each, and ‘hark back’ for some ancestor that will give us a Branching (for of course it ca’n’t give us any single Letter) that we’ve not yet used.”

“Now, to hark back. Will v do? No. Will A? Yes, it will: we’ve not used 19 yet. So of course No. 2 would be a superfluous Premiss, were it not that it happens to be used in the other Branch.”

“But stay! We’ve had both these Letters before! There they are, away on the left, supplied by No. 22, and calling $H^{\prime }$ their father! Well, these are very affectionate children: they don’t seem to mind who is to be called their father, so long as somebody will own them! Well, one of the two sets must wait, anyhow, and see what happens to the other set. Which shall it be? This new set? Well, it could only utilise the experiences of the other l and $n^{\prime }$, provided that they don’t use, in their annulment, either a or $H^{\prime }$, for those do not occur in the ancestral line of this new set. This we must look into. I see that a occurs in 10, 11, and 16. It ca’n’t use 10 again, as it used that before we got down to l and $n^{\prime }$. No. 11 it ca’n’t use, because that wants t: and No. 16 it ca’n’t use, because it wants E. Well, a is safe, then. And $H^{\prime }$ occurs only in 22, which it uses in branching. So this new set of l and $n^{\prime }$ may wait.”

“Now we go back to the $Bt$-Branch. What will $m^{\prime }$ do for us? It occurs in 3 and 9. In 3 it gives us $k^{\prime }$: in 9 it divides. And what will $E^{\prime }$ do? In 1 it divides: but in 14 it gives us $c^{\prime }$.”

“Now we return to the extreme left. What will $C^{\prime }$ do? $C^{\prime }$ occurs only in 4; but that’s very helpful, as it gives us two fresh Letters at once, $b^{\prime }$ and $n^{\prime }$.”

“Now for b. Well, b occurs in no less than four Premisses. It ca’n’t use 4, as that would want l as a partner: but it can use 5; and that gives us $s^{\prime }$. Also it can use 9 (or rather the second bit of 9); and that gives us $e^{\prime }$. No. 15 it ca’n’t use yet.”

“Now we return to the $Bt$-Branch. What can we do with $k^{\prime }$? It only occurs in 20, and that divides. Is $c^{\prime }$ of any use? Yes, in 10 it gives us $H^{\prime }$: 19 it ca’n’t use.”

“Now we return to the extreme left. What can we do with $b^{\prime }{n}^{\prime }$? Well, $b^{\prime }$ occurs in 13, along with $n^{\prime }$, and also with D, $r^{\prime }$, and z, all of which we’ve got upstairs! So here’s another Nullity!”

“Now for $s^{\prime }{e}^{\prime }$. Well, $s^{\prime }$ occurs in 15, which gives us another Nullity!”

“Come! That finishes up all the branches on this side, except the two that are waiting, l and $n^{\prime }$; and those we know are all right: we’ve discussed that matter already.”

“Now there’s nothing left but the $Bt$-Branch. What can we do with $H^{\prime }$? Can we utilise, for its benefit, the $H^{\prime }$ that has already appeared, higher up, in the left-hand Branch? I must examine the Branches dependent from the earlier $H^{\prime }$, and refer to the List of Premisses, to see whether all these, used in its annulment, can lawfully be used here.”

“No, I find that the earlier uses 13 in both the Branches dependent from it: and that requires $r^{\prime }$: and that we haven’t got here. So this $H^{\prime }$ must get annulled in some other way. What can we do with it? Well, we must divide here.”

“Now, would $l^{\prime }$ or n be of any further use? Yes, $l^{\prime }$ would.”

“Now what will l do? In 1 it gives us $C^{\prime }$: 4 it ca’n’t use yet.”

“Now for $n^{\prime }{l}^{\prime }$. What will $n^{\prime }$ do? It only occurs in 13, and there it divides. Let’s try $l^{\prime }$. In 15 it divides: 19 it can’t use—nor 22. Well, then, we must divide. Let’s do it with 13.”

“Now, would $b^{\prime }$ or $r^{\prime }$ be of any further use? No, neither of them: so there’s no tacking on to be done. Now for $C^{\prime }$. In 4 it gives us two Letters at once, $b^{\prime }$ and $n^{\prime }$.”

“Now for that Branching. What will b do? It ca’n’t use 4—nor 5, since we’ve got $s^{\prime }$ upstairs: in 9 it gives us $e^{\prime }$: and in 15 it gives us e. So we’ve finished that Branch.”

“Now for r. In 9 it gives us $e^{\prime }$: 17 it ca’n’t use yet: in 20 it gives us $h^{\prime }$: 24 it ca’n’t use.”

“Now back to the l-Branch. Our last entry was $b^{\prime }{n}^{\prime }$. What will $b^{\prime }$ do? In 13 it gives us r: that’s all it will do.”

“Now back to the extreme right. What will $e^{\prime }$ do? In 15 it gives $b^{\prime }$; but in 17 it gives us a Nullity! So we needn’t trouble about 15.”

“Now there’s nothing left but the l-Branch. Our last entry was r: and, as we’ve just annulled an r on the extreme right, we may as well utilise it, if possible. Let’s see if this new r can lawfully use 9, 20, and 17.”

“Yes, it can.”

“So now the Tree is finished! And we’ve proved $dzD$ to be a Nullity. Let’s see if any of them exist separately.”

“Yes, $dz$ exists in 24. So now for our Conclusion.”

“Now, was No. 12 superfluous, after all?”

“No, it wasn’t: we had to use that $s^{\prime }$ in order to bring in No. 15. So, ‘now my task is fairly done, I can fly or I can run’—only, I ca’n’t fly, and, on the whole, I prefer not to run!”

Here ends my long (and, I fear, tedious) soliloquy. But does not my exhausted Reader, who has patiently obeyed all its instructions, feel a certain glow of pride at having constructed so splendid a Tree—such a veritable Monarch of the Forest?

We have now completed the Solution of this Problem. But it is always desirable to verify every such Tree, by translating it into Sorites-form: this will require a supplementary soliloquy, with stage-directions as before.

“Now let’s verify this Tree. At Branching 21 I take the $t^{\prime }$-Branch first: and in it, at Branching 20, I take $r^{\prime }$ first. Under $r^{\prime }$, at Branching 7, a and A are both single Letters. Well, let’s take a first. Under a, of course I take l first: and, as that ends with a circle, we can begin with that Branch, which must be numbered 25, as there are 24 Premisses.”

“Now, which Partial Conclusion shall we take for 26? Best take the other part of Branching 22.”

“Then of course we go up this Branch for 27. The Sorites will begin with 26: then cross by bridge 22: then take in 25: then upstairs, and take in 10—and there you are!”

“Then, for 28, of course we must work up to $r^{\prime }$, just above. The Sorites will be—we must take the A-Branch first, as it isn’t yet worked up to the top—the Sorites will be 26 (we must take the $n^{\prime }$-Branch first, as it refers to the biliteral Branch $n^{\prime }{l}^{\prime }$): then cross by 19: then take in 25: then, upstairs and take in 2. Now we’ve got to A. Then cross by the 7-bridge: then take in 27: that finishes it.”

“Then, 29 must come at the top of the $t^{\prime }$-Branch. The sorites must begin with the circle at the foot of the $h^{\prime }r$-Branch. So it will be 17, 9*: then the 20-bridge: then take in 28: then upstairs, and take 3, 14, 18, and 23. That gives us 29.”

“Now for the great $Bt$-Branch. At Branching 7 of course we take a: and, under it, at Branching 22, we take l: and that ends in a circle: so let’s begin there. But we mustn’t do it all at once: a Partial Conclusion must be recorded at r, for the benefit of the r-Branch just to the right, so the Sorites will be 17, 9*, and 20.”

“Then we had better have 31 at the top of this same Branch: and the Sorites will be 30, 13, 4, and 1.”

“Well, now for the $n^{\prime }{l}^{\prime }$-Branch. It doesn’t matter which we take first, b or r: both are single Letters: but b wants working up: so of course we begin there. Our Sorites will be 9**, 15: then bridge 13: then take in 30: that brings us up to $n^{\prime }{l}^{\prime }$: then bridge 22: then take in 31: then upstairs, taking 10, 3, 14, 11, 16: then we must record, as a is the single-Letter Branch.”

“Now, there’s only one more Sorites wanted: so there’ll be no more recording to do. Our final Sorites must begin with 24, to take in the $r{a}^{\prime }$-Branch: then cross by the bridge 7: then take in 32: then—do we go up to $A^{\prime }$ at once? Or do we take in $s^{\prime }$? Oh, I remember! We are not to miss $s^{\prime }$: it’s used down below. Well, then, the Sorites goes on with 12, 2, 8, 6: then bridge 21: then take in 29: and that ought to give us our final Nullity $dz{D}_0$!”

…

Book XXI. Logical Puzzles

Chapter I. Introductory

Under this general heading I shall discuss various arguments, which are variously described by Logical writers. Some have been classified as ‘Sophisms’, that is, according to etymology, “cunning arguments”, whose characteristic Attribute seems to be that they are intended to confuse: others as ‘Paradoxes’, that is, according to etymology, “things contrary to expectation”, whose characteristic Attribute seems to be that they seem to prove what we know to be false: but all may be described by the general name “Puzzles.”

Chapter II. Classical Puzzles

§ 1. Introductory

I shall here enunciate five certain well-known Puzzles, which have come down to us from ancient times, and which the Reader will no doubt like to know by their classical titles.

§ 2. Pseudomenos

This may also be described as “Mentiens”, or “The Liar”. In its simplest form it runs thus:——

“If a man says ‘I am telling a lie’, and speaks truly, he is telling a lie, and therefore speaks falsely: but, if he speaks falsely, he is not telling a lie, and therefore speaks truly.”

§ 3. Crocodilus

That is, “The Crocodile”. This tragical story runs as follows:——

“A Crocodile had stolen a Baby off the banks of the Nile. The Mother implored him to restore her darling. “Well”, said the Crocodile, ‘if you say truly what I shall do, I will restore it: if not, I will devour it.” “You will devour it!” cried the distracted Mother. “Now”, said the wily Crocodile, “I cannot restore your Baby: for, if I do, I shall make you speak falsely: and I warned you that, if you spoke falsely, I would devour it.” “On the contrary”, said the yet wilier Mother, “you cannot devour my Baby: for, if you do, you will make me speak truly, and you promised me that, if I spoke truly, you would restore it!” (We assume, of course, that he was a Crocodile of his word; and that his sense of hnour outweighed his love of Babies.)

§ 4. Antistrephon

That is “The Retort”. This is a tale of the law-courts.

“Protagoras had agreed to train Euathius for the profession of a barrister, on the condition that half his fee should be paid at once, and that the other half should be paid, or not paid, according as Euathius should win, or lose, his first case in Court. After a time, Protagoras, becoming impatient, brought an action against his pupil, to recover the second half of his fee. It seems that Euathius decided to plead his own cause. “Now, if I win this action”, said Protagoras, “you will have to pay the money by the decision of the Court: if I lose it, you will have to pay by our agreement. Therefore, in any case, you must pay it.” “On the contrary”, retorted Euathius, “if you win this action, I shall be released from payment by our agreement: if you lose it, I shall be released by the decision of the Court. Therefore, in any case, I need not pay the money.”

§ 5. Achilles

This may be described, more fully, as “Achilles and the Tortoise”. The legend runs as follows:——

Achilles and the Tortoise were to run a race on a circular course; and, as it was known that Achilles could run ten times as fast as the Tortoise, the latter was allowed 100 yards’ start. There was no winning-post, but the race was to go on until Achilles either overtook the Tortoise or resigned the contest. Now it is evident that, by the time Achilles had run the 100 yards, the Tortoise would have got 10 yards further; and so on for ever. Hence, in order to overtake the Tortoise, he must pass over an infinite number of successive distances. Hence, Achilles can never overtake the Tortoise.

§ 6. Raw Meat

Chapter III. Other Puzzles

§ 1. About Less

§ 2. Men Tall and Numerous

§ 3. The Socialist Orator and the Irish Mob

“Isn’t one man as good as another?” demanded a Socialist orator, addressing an Irish mob. “Av coorse he is”, was the eager response, “and a great deal betther!”

§ 4. Death at Any Moment

“You may die at any moment, and probably will.” (See The Mystery of Ms. E. Drood, by Orpheus C. Kerr, p. 136/217.)

The first part of this statement seems reasonable enough: the second is obviously absurd. Yet how can it be absurd to assert that you will do what it is quite reasonable to assert that you may do?

§ 5. The Small Girl and Her Sympathetic Friend

Small Girl: I’m so glad I don’t like asparagus!

Sympathetic Friend: Why, my dear?

Small Girl: Because, if I did, I should have to eat it—and I ca’n’t bear it!

Examine the reasoning process, if any, which has taken place in the mind of the Small Girl.

A Notice at the Seaside

The blue ensign denotes that all boats, licensed to carry from two to four persons, are prohibited from putting off, and no other boat must put to sea without a licensed boatman.

The red ensign denotes that only large rowing-boats carrying from five to seven persons, and sailing boats, can put to sea with boatmen.

[N.B. The above Notice is exhibited at a certain seaside-place not 100 miles from the Needles.]

What boats, if any, can put to sea when one of these two ensigns is hoisted, but cannot when the other is hoisted?

Chapter IV. Solutions of Classical Puzzles

§ 1. Introductory

The following Solutions have not come down to us from ancient times, but are merely modern speculations, which the Reader can take or reject “at his own sweet will.”

§ 2. Pseudomenos

This Puzzle might be described as a “Paradox”, since it seems to prove that the man in question is speaking both truly and falsely at the same moment.

The best way out of the difficulty seems to be to raise the question whether the Proposition “I am telling a lie” can reasonably be supposed to refer to itself as its own subject-matter: to which the answer seems to be that it can not, since its doing so would lead to an absurdity.

Symbolically, it may be solved as follows:

Let Univ. be “Cosmophases”; $a=$ the man speaks truly; $b=$ the Proposition, stated by the man, can be its own subject-matter.

Then we have $a⁋a^{\prime }b$, and $a^{\prime }b⁋a$; i. e. $ab{a}_0†a^{\prime }b{a}_0^{\prime }$; i. e. $a{b}_0†a^{\prime }{b}_0$, which together prove $b_0$: i. e. the Proposition can not be its own subject-matter.

Let us now consider what result would have followed if the man’s statement had been “I am telling the truth” instead of “I am telling a lie.”

Here, if we assume that the Proposition can be its own subject-matter, it evidently follows that, if he speaks truly, he is telling the thruth, and that, if he speaks falsely, he is not telling the truth. In either case, the supposition does not lead to any absurdity.

Symbolically, it would run as follows, using the same Dictionary as before: $ab⁋a$, and $a^{\prime }b⁋a^{\prime }$; i. e. $ab{a}_0^{\prime }†a^{\prime }b{a}_0$; each of which Nullities is a truism, and proves nothing as to the existence, or non-existence, of b. Hence there is nothing in the Data to prevent the Proposition from being its own subject-matter, and nothing to tell us whether the man is speaking truly or falsely.

In short, such Data lends to no result at all, and it is not worth while to discuss them.

This Puzzle is sometimes given in a more complex form, viz.

Epimenides the Cretan asserted that Cretans were always liars. If he spoke truly, they were always liars, and therefore he, being a Cretan, was lying at that moment, and therefore spoke falsely. Hence he did not speak truly; that is, he must have been lying.

Here we cannot show that, if he spoke falsely, he also spoke truly. Hence no absurdity follows from the supposition that the Proposition can be its own subject-matter; and the conclusion, however unwelcome to Epimenides, is correct.

Symbolically, it runs thus:

Let Univ. be “Cosmophases”; $a=$ Epimenides speaks truly; $b=$ the Proposition, stated by him, can be its own subject-matter.

Then we have $ab⁋a^{\prime }$; i. e. $ab{a}_0$; i. e. $a{b}_0$; i. e. $b⁋a^{\prime }$; i. e. On the supposition that the Proposition, stated by Epimenides, could be its own subject-matter, he was certainly lying.

If the statement made by Epimenides had been “The Cretans always speak the truth,” there would have been nothing in the Data to prevent the Proposition from being its own subject-matter, and nothing to tell us whether Epimenides was speaking truly or falsely.

§ 3. Crocodilus

On this Sophism Lotze makes the discouraging remark, “There is no way out of this dilemma.” I think, however, that we shall find the machinery of Symbolic Logic sufficient for its solution.

Let Univ. be “Cosmophases”; $a=$ the Mother speaks truly; $b=$ the Crocodile keeps his word; $c=$ the Crocodile devours the Baby.

Then we have, as Data, $$\begin{array}{ccccccccccc}1& & 2& & 3& & 4& & 5& & 6\\ a{b}_1{c}_0& †& a{b}_1^{\prime }{c}_0^{\prime }& †& a^{\prime }{b}_1{c}_0^{\prime }& †& a^{\prime }{b}_1^{\prime }{c}_0& †& c_1{a}_0^{\prime }& †& c_1^{\prime }{a}_0\end{array}$$

Here we may ignore 2, 4, as being contained in 6, 5; and we see, by inspection, that b is the only Retinend.

$\therefore b_0$; i. e. Whatever the Crocodile does, he breaks his word.

Thus, if he devours the Baby, he makes her speak truly, and so breaks his word; and if he restores it, he makes her speak falsely, and so breaks his word. His sense of honour being thus hopeless of satisfaction, we cannot doubt that he would act in accordance with his second ruling passion, his love of Babies!

§ 4. Antistrephon

The best way out of this Paradox must seem to be to demand an answer to the question “Which of the two things, the agreement and the decision of the Court, is to over-rule the other, in case they should come into collision?”

The Data do not enable us to answer this question. Protagoras naturally makes one, or the other, supreme, as best suits his purpose: and his docile pupil follows his example.

The right decision of the Court would obviously be against Protagoras, seeing that the terms of the agreement were still unfulfilled. And, when that decision had been pronounced, the practical result would be that, if the agreement was to be supreme, Euathius would have to pay the money: if the decision of the Court was to be supreme, he would be released from payment.

§ 5. Achilles

This is a mathematical Fallacy, and involves the false assumption that a series of distances, infinite as to number, is also infinite as to total length.

Here the assumption is that $$\left(111+\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+\text{ \&c.}\right)$$ of a mile, where the number of terms can be made greater than any assigned number, can be made greater than any assigned length. But the above series is the circulating decimal 111.1 which as the Reader probably knows, can never reach the limit $111\frac{1}{9}$. Hence, by the time Achilles has run $111\frac{1}{9}$ yards, he must necessarily have overtaken the Tortoise.

§ 6. Raw Meat

The best way of escaping from this savage Paradox seems to be to introduce the Dated Copula. Two epoches have to be taken into consideration, viz.

At epoche (a), the second Premiss is true; and the piece of meat in question possesses, at that moment, the Pair of Attributes, “bought by me in the market” and “raw.”

At epoche (b), the first Premiss becomes true, and the second ceases to be true. That is to say, the piece of meat possesses, at that moment, the Pair of Attributes, “eaten by me at dinner” and “bought by me in the market,” but it has ceased to possess the Attribute “raw”.

Hence, we cannot assert that the two Premisses are true at the same moment.

Hence there is no Conclusion.

Chapter V. Solutions of Other Puzzles

§ 1. About Less

This (apparently valid) Syllogism belongs to the Class “Paradoxes,” since it seems to prove that 5 is less than 10 and more than 6.

The first thing to be said about it is that its second Premiss is mere tautology, being of the form “All $xy$ are x.” Hence, if the Conclusion follows at all, it must follow as an Immediate Inference from the first Premiss.

But such an Inference would involve the Fallacy of Vox Ambigua, since the phrase “speaks truly” is capable of two interpretations, viz.

With (a), the first Premiss cannot be accepted as true, unless we are assured that the speaker says no more.

With (b), the Inference is a valid one, and the Conclusion true.

§ 2. Men Tall and Numerous

This involves the Fallacy of Vox Ambigua. The phrase “men over 5 feet high” may be taken to mean, either “every man over 5 feet high,” or “the Class composed of men over 5 feet high” regarded as one single Thing. (See Part I, Book I, Chapter II.)

With the first interpretation, the Premisses are not true: the Attribute “numerous” cannot be applied to an individual man.

With the second interpretation, the Subject of each Premiss is a single Thing: and what the Conclusion asserts is that one of these two single Things in not the other. In this case, the Syllogism is valid, and the Conclusion is true.

§ 3. The Socialist Orator and the Irish Mob

At first sight, this reply might be thought to support the position taken by the orator: but, on further examination, it is seen to contradict it. The Paradox, here involved, may be logically stated as follows:

The orator’s implied assertion is that, in every Pair of men, each is not less good than the other: from which it may easily be proved that every Pair of men possesses the Attribute “equal in merit.”

Pat’s ready reply asserts that, in every Pair of men, each is better than the other; i. e. that every Pair of men possesses the Attribute “unequal in merit.” This not only contradicts the previous assertion, but also contradicts itself, since it may easily be shown to involve the assertion that each is at once better, and worse, than the other. But self-contradiction, in a Proposition, is not an Attribute that would for a moment discredit it in the Emerald Isle!

§ 4. Death at Any Moment

The best explanation I can find, for this bewildering Paradox, does not altogether satisfy me; and I shall be grateful to any Reader who will suggest a better. My solution is as follows:

This is a mathematical Paradox. The first clause of it asserts that, at any given moment during a certain period (say the next ten years), the death of Mr. E. Drood possibly may occur: Now, if we take n to represent the number of moments in the period, and assume that the event is equally likely at each moment, the probability of its occurrence is $1/n$th of certainty, and therefore is not (what the word “probable” usually implies) greater than one-half of certainty. Hence the Proposition is necessarily false, even if we assign to n the minimum possible value, viz. 2. If, however, we re-word the Proposition thus, “At any given moment, during the next period of n moments, your death possibly may occur, and there is a probability, amounting to $1/n$th of certainty, that it will occur,” we make it logically correct. But it is to be feared that it has lost, during the corrective process, all the sparkle and humour with which it came from the pen of its ingenious author!

Book XXII. Solutions of Problems Set by Other Writers

Chapter I. Problems

The books, from which the following twenty-five Problems are quoted, are as follows:

1. [F] Vol. I, p. 77

Let $a=$ planets; $b=$ fixed stars; $c=$ subject to gravity. The Reader will try in vain to produce from these Premisses, by legitimate substitution, any relation between b and c.

2. [B] p. 124. Quoted in [H] p. 432

Every A is one only of the two B or C. Every D is both B and C, except when B is C, and then it is neither.

3. [J] p. 342

If x that is not-a is the same as b, and a that is not-x is the same as c, what is x in terms of a, b, and c?

4. [J] p. 340

At a certain town where an examination is held, it is known that

Show that, if this be so, there can be no candidates.

5. [F] Vol. I, p. 191

For every man in the house there is a person who is aged: some of the men are not aged.

6. [J] p. 336. Quoted in [H] p. 433

There is a certain Class of Things, from which A picks out all the x that are z and all the y that are not-z; and B picks out from the remainder the z which are y and the x that are not-y. It is then found that what is left exactly comprises the Class of z that are not-x. What can be determined as to the original Class?

7. [J] p. 350. Quoted in [H] p. 437, and [K] p. 53

Given $xy=a$, $yz=c$: find $xz$ in terms of a and c.