Symbolic Logic

by Lewis Carroll

which yield the Conclusion xy₀

We see that, in order to get this Conclusion, we must eliminate m and m′, and write x and y together in one expression.

Now, if we agree to mark m and m′ as eliminated, and to read the two expressions together, as if they were written in one, the two Premisses will then exactly represent the Conclusion, and we need not write it out separately.

Let us agree to mark the eliminated letters by underscoring them, putting a single score under the first, and a double one under the second.

The two Premisses now become

xm₀ † ym′₀

which we read as “xy₀”.

In copying out the Premisses for underscoring, it will be convenient to omit all subscripts. As to the “0’s” we may always suppose them written, and, as to the “1’s”, we are not concerned to know which Terms are asserted to exist, except those which appear in the Complete Conclusion; and for them it will be easy enough to refer to the original list.

[pg092][I will now go through the process of solving, by this method, the example worked in [§ 2.]

The Data are

1k₁l′₀ † 2dh′₀ † 3a₁c₀ † 4b₁e′₀ † 5k′h₀ † 6b′l₀ † 7d′₁c′₀

The Reader should take a piece of paper, and write out this solution for himself. The first line will consist of the above Data; the second must be composed, bit by bit, according to the following directions.

We begin by writing down the first Premiss, with its numeral over it, but omitting the subscripts.

We have now to find a Premiss which can be combined with this, i.e., a Premiss containing either k′ or l. The first we find is No. 5; and this we tack on, with a †.

To get the Conclusion from these, k and k′ must be eliminated, and what remains must be taken as one expression. So we underscore them, putting a single score under k, and a double one under k′. The result we read as l′h.

We must now find a Premiss containing either l or h′. Looking along the row, we fix on No. 2, and tack it on.

Now these 3 Nullities are really equivalent to (l′h † dh′), in which h and h′ must be eliminated, and what remains taken as one expression. So we underscore them. The result reads as l′d.

We now want a Premiss containing l or d′. No. 6 will do.

These 4 Nullities are really equivalent to (l′d † b′l). So we underscore l′ and l. The result reads as db′.

We now want a Premiss containing d′ or b. No. 4 will do.

Here we underscore b′ and b. The result reads as de′.

We now want a Premiss containing d′ or e. No. 7 will do.

Here we underscore d and d′. The result reads as c′e′.

We now want a Premiss containing c or e. No. 3 will do—in fact must do, as it is the only one left.

Here we underscore c′ and c; and, as the whole thing now reads as e′a, we tack on e′a₀ as the Conclusion, with a ¶.

We now look along the row of Data, to see whether e′ or a has been given as existent. We find that a has been so given in No. 3. So we add this fact to the Conclusion, which now stands as ¶ e′a₀ † a₁, i.e. ¶ a₁e′₀; i.e. “All a are e.”

If the Reader has faithfully obeyed the above directions, his written solution will now stand as follows:—

1k₁l′₀ † 2dh′₀ † 3a₁c₀ † 4b₁e′₀ † 5k′h₀ † 6b′l₀ † 7d′₁c′₀

1kl′ † 5k′h † 2dh′ † 6b′l † 4be′ † 7d′c′ † 3ac ¶ e′a₀ † a₁ i.e. ¶ a₁e′₀;

i.e. “All a are e.”

[pg093]The Reader should now take a second piece of paper, and copy the Data only, and try to work out the solution for himself, beginning with some other Premiss.

If he fails to bring out the Conclusion a₁e′₀, I would advise him to take a third piece of paper, and begin again!]

I will now work out, in its briefest form, a Sorites of 5 Premisses, to serve as a model for the Reader to imitate in working examples.