The Math Underground: If This Sentence is True, This Article Sucks

If This Sentence is True, This Article Sucks
Saturday June 3, 2006 by Patrick N. R. Julius

Consider this sentence:
If this sentence is true, then this article sucks.

If it’s true, then this article sucks.
But that’s what it says!
Therefore it’s true.
Therefore this article sucks.

But then, consider this sentence:
If this sentence is true, then this article does not suck.

Therefore this article does not suck!

So does it, or doesn’t it?

You could prove anything at all!
A paradox! A total explosion! Logic becomes useless!

How can we resolve this?

Well, we could respond, “Why does just saying a sentence make it true? That’s not logical!”
But as this sentence is structured, the logic is valid!

Formally, it looks like this.
Construct this statement:
$\mbox{Let } X = (X \Rightarrow Y)$
Obviously this is true:
$X \Rightarrow X$
We can substitute in:
$X \Rightarrow (X \Rightarrow Y)$
We can contract the hypotheses:
$X \Rightarrow Y$
We can substitute again:

So we’ve proved that this is true:
$X \wedge X \Rightarrow Y$
And obviously this follows:

It doesn’t matter what Y is! No matter what, it must be true!

But obviously we can’t let this happen, can we? If we do, then we can prove all sorts of contradictions, indeed prove anything at all, and make logic meaningless!

So which step can’t we do? We’ll work backward.
$(X \wedge X \Rightarrow Y) \Rightarrow Y$
In words, “X is true and X implies Y means that Y is true.”
No, that has to be true; otherwise implication means nothing at all.

$(X \Rightarrow Y) \Rightarrow X$
“If X implies Y, then X is true.” Well, that seems odd, but it follows from our definition of X.

$(X \Rightarrow (X \Rightarrow Y)) \Rightarrow (X \Rightarrow Y)$
“If X implies that X implies Y, then X must imply Y.” How could this not be true? When X is true, X implies Y, and so Y is true. When X is true, Y is true. Hence X implies Y.

$(X \Rightarrow X) \Rightarrow (X \Rightarrow (X \Rightarrow Y))$
Really odd-looking statement, but yes, it must be true if and $X \Rightarrow Y$ are the same thing.

We’re running out of statements!
$X \Rightarrow X$
“X is true implies X is true.”
That’s what we call a… tautology. It has to be true.

So there’s only one statement left, and no obvious reason that it should be false:
$\mbox{Let } X = X \Rightarrow Y$

Sure, it seems oddly self-referential, but all we’ve done is defined a statement in terms of other statements; what’s wrong with that?

Indeed, in material, strict, and relevant implication, this construction is completely proper and this proof is completely valid.

So should we abandon logic altogether?
Hardly.

Instead we should realize the deeper reason why this first assumption is total nonsense.

If I said to you, “Let X be the frog which is truth,” or “Let Q be the number which is a planet,” or “Let M be the coffee cup which is a philosopher,” you’d think I was crazy.
What I’ve committed here is called a category error. I’ve asked you to examine the relationship between two entities which are fundamentally incomparable.
Colloquially, I’m comparing apples and oranges.

In making the same statement $\mbox{Let } X = (X \Rightarrow Y)$ , I’ve committed exactly the same error.

Let me put it this way:
There are empirical statements, tautological statements, and hypothetical statements.
“The Earth revolves around the Sun” is an empirical statement. It’s either true or it’s not, and can be determined as such by physical observation.
“1 + 1 = 2” is a tautological statement. It’s either true or false by definition, and couldn’t possibly be any other way.
“ $n \in N \Rightarrow n + 1 \in N$ ” doesn’t fall into either of these categories. It’s a different animal, a hypothetical statement. It’s “true” because its implication is valid. It doesn’t need its premise to be true or its conclusion to be true, only their relationship.
Indeed, it’s what I’d call a first-order hypothetical statement, because it is made up of implications between empirical statements.
$(n \in N \Rightarrow n + 1 \in N) \Rightarrow (1 \in N \Rightarrow 2 \in N)$ would then be a second-order hypothetical statement, because it is made up of implications between first-order hypothetical statements.

In this way you can define the various levels as nth-order hypothetical statements, where n is any natural number.

How does this help? Well, no n-order hypothetical statement could ever be equivalent to an n+1-order hypothetical statement.

When I say, “if this statement is true,” which statement am I referring to? The hypothetical statement, “if …, then this article sucks”, or the empirical statement, ”... is true”? They’re not the same thing. They can’t possibly be. So the fact that the whole hypothetical statement is true does not mean that the interior empirical statement is true.

In the same way, the concept of $X = (X \Rightarrow Y)$ becomes completely nonsensical. If is empirical (or tautological), $X \Rightarrow Y$ is 1-hypothetical. If is 1-hypothetical, $X \Rightarrow Y$ is 2-hypothetical. And on and on, always a step behind.
The two statements can never be of the same hypothetical order, and therefore can never be equivalent.

The paradox was in our first assumption, that such a statement could even exist.

Actually, I’ve been reading up a bit about that paradox, and it seems that it’s the Russell paradox of intuitivist logic. Nobody has found a satisfactory answer for it (yet), and someone is actually creating a new system of logic just to avoid it (I havn’t actually read about what his system is, but I’m planning to do it when I have the time. Who knows, I might even write about it). Personally, I find it kind of cool that it’s so easy to get to the point in which you’re discussing issues that are (almost?) at the spearhead of mathematics.

— Noam Samuel Jun 5, 07:18 AM #
What? You’re saying my answer isn’t satisfactory?

— Patrick Jun 5, 04:36 PM #
Actually, I found an even better way.

You don’t need to invoke the difference between hypothetical orders (though you can if you wish).

All you have to ask is, “when I construct a statement, have I made that statement well-defined?”

I’ll use arithmetic to demonstrate.
If I say, $\mbox{Let } x = \sqrt{x}$ , that’s well-defined. There is a non-empty set of complex numbers (namely $\{ 0, 1 \}$ ) which satisfies this requirement.
So when I say $(x = \sqrt{x}) \Rightarrow (x^2 = x)$ , that’s perfectly true.

But if I say, $\mbox{Let } x = x + 1$ , that’s not well-defined. There are no numbers $x \in C$ such that this statement is true.
So to say that $(x = x+1) \Rightarrow (0 = 1)$ is basically meaningless.

In the same way, if I say $\mbox{Let } X = X \Rightarrow Y$ , there are no statements X satisfying this criterion. The natural-language version is just a play on words, muddling what the words “this statement” mean. The set of statements satisfying this definition is empty, and so X is not well-defined. Therefore you can’t use it in any following logical statements.

— Patrick Jun 6, 12:38 PM #
By the way, I have yet another problem with the axiom of infinity.

So it’s possible for a set X to be its own cardinality?
$| X | = X$ ?

If not, we have a problem, since:
$|3| = |\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\} | = 3$ , and indeed $|n| = n , \, \forall n \in N$ .

But if it is true, then we have some very interesting infinite regressions, don’t we?
$|1| = 1 = \{\emptyset\}$
$| |1 | | = |1 | = 1$
$| | | 1 | | | = | | 1 | | = |1| = 1 \ldots$

— Patrick Jun 6, 12:46 PM #