Three of these statements are false.
  1. 2 + 2 = 4
  2. 3 * 6 = 17
  3. 8 / 4 = 2
  4. 13 - 6 = 5
  5. 5 + 4 = 9 Solution: Statements 2 and 4 are false. This makes the assertion of three false statements, itself, false. Its own falseness makes it true...

    A Blather of Paradoxes
"This statement is false."

You might be tempted to alleviate the paradox by claiming that the statement has no truth value at all (is neither true nor false). But consider this statement:

"This statement is false or it has no truth value."

If it is true, we have a contradiction.
If it is false, then it is true.
And if it has no truth value, it is also true.

There you have it. Math is inconsistent.

A = C false
B = C hasnotruthvalue
C = A v B
C true => A false & B false => (A v B) false => C false =><=
C false => A true & B false => (A v B) true => C true =><=
C hasnotruthvalue => A false & B true => (A v B) true => C true =><=

I don't see how A has to be N. Perhaps you are confusing A to be "A is false"? I think the problem here, instead, is not the misuse of the hasnotruthvalue predicate, but that our notion of "true" is not well-defined.

Bzzt, wrong, thank you for playing.

A schema in which all statements have truth values has this sort of algebra:
T V T = T
T V F = T
F V T = T
F V F = F

Now let N = the truth value "has no truth value". Let A be "this statement is false" and let B be "this statement has no truth value". So jliszka is looking to establish the truth value of A V B. We now have nine possibilities (T V T, T V F, T V N, N V F, etc) for the combinations of A and B, of which only four are covered in the truth table above.

And how do we decide the values of the others? Hmm...

Let C = (A V B). Suppose (A V B) has no truth value. Then A says C is false, and B says C hasnotruthvalue (let's make that a single predicate). So by assumption "C is false" isn't true: we can't assign the truth value "false" to C. And so...? And so C is either T or N. T would be a contradiction, but N is still consistent.

That was considering the A branch: we seem to have established that value(C) = N is the only consistent interpretation. Fine. Now let's look at the B disjunct. This says "C hasnotruthvalue". This is either T, F, or N. Well T is fine. T is consistent with what we know about A. F is bad, because it contradicts what we've just learnt about A. What about the third possibility, value(B) = N? This is claiming "'This sentence has no truth value' has no truth value" -- err, at this stage I can't make sense of that, so let's say okay, N isn't ruled out.

So A has to be N. B can be T or, possibly, N. Therefore the compound claim C = (A V B) has two possible truth values, either N V T or N V N. And those algebraically work out to...?

The fallacy is this. "X hasnotruthvalue" is true if X hasnotruthvalue. But "X hasnotruthvalue OR Y something" cannot imply anything straightforward, because the boolean algebra of OR only applies to true and false. It doesn't apply to values that are by definition outside the rules of the game. You can't invoke meta-rules which say "oh, by the way, I'm going to keep these in play for as long as it suits me, even though I agreed to exclude them".

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