If □P, does it follow that P is a tautology?
I know in K modal logic, the law of NEC states
⊢ P; therefore □P.
The corresponding conditional of the previous argument is
If ⊢ P then □P.
Now ⊢P iff P is a tautology, where I, as a monist, have concluded the propositional calculus is the only logic that makes sense.
So if that conclusion of mine is correct, then by the rule of replacement
If P is a tautology, then □P.
Can anyone out there can prove the converse of the preceding statement?
EDIT -
Let Rx be the statement x is red all over, and let Bx be the statement x is blue all over. If Rx then not Bx. From the definition of IF, not Rx or not Bx. Now by DeMorgan's Law not(Rx & Bx). The previous statement is not a tautology unless Bx iff not Rx. Since there's more than two colors, it's not the case that Bx iff not Rx. So not(Bx and Rx) is not a theorem of any formal logistic system that is analytic. It is a statement about reality, that denotes a proposition that is true at all moments in time. The proposition it denotes has a truth value that is constant in time, so only one row of its truth table is possible, for any specific object x. Consider the following truth table, for an arbitrary object x.
Rx | Bx | not(Rx and Bx) |
---|---|---|
0 | 0 | 1 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
The fourth state row is impossible. As you can see, the statement in column three isn't a tautology, unless we carefully define tautology so that it is. As only three rows are possible we may say
(not Rx and not Bx) or (not Rx and Bx) or (Rx and not Bx), for any object x.
Thus if we define a tautology to be a compound statement that is true in every row of its truth table that is possible, then the statement not(Rx and Bx) is a tautology.
Can we redefine tautology in this manner?