The (almost really) Complete Works of Lewis Carroll

There are three Propositions, A, B, and C.

It is given that
“If A is true, B is true. (i)
If C is true, then if A is true B is not true. (ii)

Nemo and Outis differ about the truth of C.

Nemo says C cannot be true: Outis says it may be.

Nemo’s Argument

Number (ii) amounts to this:—

If C is true, then (i) is not true.

But, ex hypothesi, (i) is true.

∴ C cannot be true; for the assumption of C involves an absurdity.

Outis’s Reply

Nemo’s two assertions, “if C is true, then (i) is not true” and “the assumption of C involves an absurdity”, are erroneous.

The assumption of C alone does not involve any absurdity, since the two Hypotheticals, “if A is true, B is true” and “if A is true, B is not true”, are compatible; i. e. they can be true together, in which case A cannot be true.

But the assumption of C and A together does involve an absurdity; since the two Propositions, “B is true” and “B is not true”, are incompatible.

Hence it follows, not that C, taken by itself, cannot be true, but that C and A cannot be true together.

Nemo’s Rejoinder

Outis has wrongly divided Protasis and Apodosis in (ii).

The absurdity is not the last clause of (ii), “B is not true”, but all that follows the word “then”, i. e. the Hypothetical “If A is true B is not true”; and, by (ii), it is the assumption of C only which causes this absurdity.

In fact, Outis has made (ii) equivalent to “If C is true [and if A is true] then if A is true B is not true”. This is erroneous: the words in the brackets in the compound Protasis are superfluous, and the remainder is the true Protasis which conditions the absurd Apodosis, as is evident from the form of (ii) originally given.

This Theorem in Hypotheticals—that the Propositions, numbered (i) and (ii), together prove that C cannot be true—may be illustrated by the following algebraical example:—

Let $ax+(a-b)y+z=5$; (1)
$bx+z=6$ (2)

Equation (1) may be stated as a Hypothetical, thus:—

“If $ax$, $(a-b)y$, and z be added together, the number ‘5’ is obtained”.

Let ‘A’ mean “$ax$, $(a-b)y$, and z are added together”;
‘B’ ” “the number ‘5’ is obtained”;
‘C’ ” “$a=b$”.

Then we have
“If A is true B is true”.

Assume that C is true; i. e. that $a=b$.

Then $(ax+(a-b)y+z)$ becomes $(bx+z)$, which, by Equation (2), must always $=6$.

Hence
“If C is true, then if A is true B is not true”.

Therefore C cannot be true;
i. e. ‘a’ cannot = ‘b’.

Outis’s Second Reply

This reply will include (α) a proof that “Nemo’s Argument” is self-destructive; (β) a proof that his algebraical example fails, owing to its not correctly representing the data; (γ) a proof that, when corrected, it illustrates Outis’s contention, viz. that Hypotheticals (i) and (ii) prove, not that C, taken by itself, cannot be true, but that C and A cannot be true together; (δ) a simple proof of the true outcome of these two Hypotheticals.

(α)

Let us consider the Trio of Hypotheticals (which we will call (K), (L), and (M))

It will not be disputed that (L) and (M), taken together, are equivalent to the Categorial (which we will call ‘N’) “Y is true.” Hence the above Trio of Hypotheticals is equivalent to the Hypothetical and Categorial

For this Trio (or its equivalent Pair) two different interpretations might be proposed, viz.

These interpretations are incompatible.

Now, when Nemo says “the assumption of C involves an absurdity”, the “absurdity”, to which he alludes, is the simultaneous truth of the two Propositions “If A is true B is true” and “If A is true B is not true”.

These two Propositions are Hypotheticals of the forms (L) and (K): and, in declaring that the assumption of their simultaneous truth involves an absurdity, Nemo virtually declares that they cannot be true together.

Here, then, he adopts the first interpretation of the Trio of Hypotheticals, (K), (L), and (M).

Again, when he says “∴ C cannot be be true”, the premisses, from which he deduces this conclusion, are the two Propositions “If C is true, then (i) is not true. But, ex hypothesi, (i) is true”.

These two Propositions are a Hypothetical and a Categorial of the forms (K) and (N): and, in deducing a conclusion from them, regarded as premisses, Nemo virtually declares that they can be true together.

Here, then, he adopts the second interpretation of the Trio of Hypotheticals, (K), (L), and (M).

Thus, he has adopted, in the course of one and the same argument, two incompatible interpretations of this Trio.

Hence, “Nemo’s Argument” is self-destructive.

(β)

Let us now examine Nemo’s algebraical example.

He gives us Equations (1) and (2) as always true.

Hence Equation (1) remains true, even when $a=b$.

Hence his second Hypothetical is incomplete: it ought to be “If C is true, then if A is true B is (by Equation 1) true, but (by Equation 2) not true.”

Hence his algebraical example fails, owing to its not correctly representing the data.

(γ)

The two Hypotheticals, when fully stated, run thus:—

These two may be stated as three Hypotheticals, viz.

The second of these we may omit, as it leads to no result. The other two may be more briefly stated thus:—

And the correct conclusion is, not that C, taken by itself, cannot be true, but that C, A, (1), and (2) cannot all be true together.

But A is always possible; so that we may, if we like, assume it is always true, and not mention it.

The two Hypotheticals may now be written thus:—

Therefore C and (1) and (2) cannot all be true together, though any two of them may be true by themselves.

Thus, if C and (1) are true, then (2) cannot be true: that is, if $a=b$ (so that Equation (1) becomes “$bx+z=5$”), and if Equation 1 is true, and if $bx+z=6$, then it cannot be true that $$ax+(a-b)y+z=5\text{.}$$

Thirdly, if (1) and (2) are true, then C cannot be true: that is, if both the given Equations are true, then a cannot $=b$.

This algebraical example might easily mislead an unwary reader, from the fact that its Conclusion, “C cannot be true,” is (on the assumption that Equations 1 and 2 are always true) a true one. The fallacy lies in prefixing the word “Therefore,” and thereby asserting that this Conclusion follows from the two Hypotheticals. This is not the case: the real reason, why C cannot be true, is that it is incompatible with Equations 1 and 2 (by subtraction we get $(a-b)(x+y)=-1$, whence it follows that $(a-b)$ cannot $=0$; i. e. that a cannot $=b$: the two Hypotheticals, by themselves, do not prove it.

(δ)

The true outcome, of the original Hypotheticals numbered (i) and (ii), may be very simply exhibited as follows:—

Let ‘t’ stand for “true”, and ‘f’ for “false”.

There are 8 conceivable combinations of A, B, and C, with regard to truth and falsity: these are as follows:—

Of these, Nos. 3 and 4 are forbidden by (i), and No. 1 is forbidden by (ii).

The other 5 combinations are possible; and two of them, viz. Nos. 5 and 7, contain the condition “C is true”, which Nemo believes to be impossible.