Originally posted by twhitehead
I did.
[b]Your argument would not show this.
Yes it did. If it did not, then explain why not.
Your argument is basically that one can know that a proposition P has a determinate truth value and that its truth value is unknown.
In what way does that differ? If English is significantly different to logic-speak in this case then please tr ...[text shortened]... conclusion is false. Once that is done, you or DeepThought can sus out the flaw in the argument.[/b]
You claim to have provided a counterexample to the idea that propositions of the form “P and P is unknown“ represent an unknowable class. So you need to provide an example where a proposition of that form is known. Your argument, again, is that one can know that a proposition has a determinate truth value and that this truth value is unknown. Fine, let’s go through it in steps.
That a proposition P has a determinate truth value translates in logic to (P or ~P). And that this truth value is unknown would translate to (~K(P) & ~K(~P)). So, your argument if successful shows:
K[(P OR ~P) & (~K(P) & ~K(~P))]. (1)
Now, we can ask ourselves: does this provide the counterexample you claim it does?
First, we can check if it simply directly satisfies. So let (P OR ~P) = Q. Does (1) have the form K[Q & ~K(Q)]? Well, it would if it happened to be the case that ~K(P OR ~P) were logically equivalent to (~K(P) & ~K(~P)). That would require ~K(P OR ~P) iff (~K(P) & ~K(~P)). But that clearly does not hold. The <-- direction is particularly problematic: just from the fact that individual disjuncts are not known, it does not follow that their disjunction is not known. So, your example does not directly satisfy as a counterexample of the right type.
Second, we can check if a counterexample of the right type follows from (1) according to valid logic and analytic operations (and let us assume the axiom about knowledge being closed under entailment to give you the best benefit of the doubt). Just consider all of the logical combinations that result from (P OR ~P) & (~K(P) & ~K(~P)). This is logically equivalent to the following:
[P & ~K(P) & ~K(~P)] OR [~P & ~K(P) & ~K(~P)] (2)
But we know some of the atomic components are analytically related. In particular, if P is true, then ~K(~P) follows analytically and if ~K(~P) is false, then it follows analytically that P is false, so P and ~K(~P) rise and fall together; likewise if ~P is true, then ~K(P) follows analytically and if ~K(P) is false, then it follows analytically that ~P is false, so these rise and fall together as well. So (2) can be simplified accordingly, and at the end of the day your (1) is equivalent to:
K[(P & ~K(P)) OR (~P & ~K(~P))]. (3)
Now, the problem is that there is no valid way to get from (3) to the counterexample you want. As DT already correctly pointed out in his assessment of your argument, the knowledge operator does not distribute over a disjunction. So, you cannot make a step here from knowledge of a disjunction to knowledge of an individual disjunct.
So, your argument fails to provide the counterexample that you claim it does. In summary, again, it fails to provide a direct counterexample because from the fact that individual disjuncts are not known, it does not follow that their disjunction is not known. It also fails to entail a counterexample of the right type under further operations because, as DT already pointed out, knowledge does not distribute over a disjunction in the manner your argument would need in order to be successful.
Your argument, if successful, would show that at least one church proposition is true. But it does not show that one is known.