allzermalmer

Truth suffers from too much analysis

Knowability Paradox and Modal Realism

Posted by allzermalmer on August 17, 2013

Equivalency is defined this way: (p–>q)<–>(q–>p)

So if we assume (p–>q) is true and (q–>p) is true, then it necessarily follows that (p–>q) if and only if (q–>p).

(1) p–><>p
(2) <>p–>p [Modal Realism]
(3) (p–><>p)<–>(<>p–>p)

There is one thing that must be made clear.

(1) is necessarily true. It is not possible that it is not true. It is an axiom of modal logic. Now (2) isn’t necessarily true. It is possible that it is not true. It is not an axiom or theorem of modal logic.

(1) can be substituted with variable of x. So x if and only if p–><>p. (1) being necessarily true implies necessarily x. (1) and x, are analytically true.

(2) can be substituted with variable of y. So y if and only if <>p–>p. (2) being not necessarily true implies possibly not y. (2) or y, are contingently true.

(3), based on substitutions of (1) & (2), takes on the form of x<–>y. Or we can say analytically true if and only if contingently true.

x is analytic implies either necessarily x or necessarily not x. [](p–><>p) v []~(p–><>p).
y is contingent implies possibly y and possibly not y. <>(<>p–>p) & <>~(<>p–>p).

(3) shows that we have collapsed any modal distinction between possibility and actuality. There is no modal difference between possibly true and actually true. This specific proposition presents that possible if and only if actual.

(4) p–>Kp [Fitch’s Theorem]
(5) Kp–>p
(6) (p–>Kp)<–>(Kp–>p)

There is one thing that must be made clear.

(5) is necessarily true. It is not possible that it is not true. It is an axiom of epistemic logic. Now (4) is also necessarily. It is not possible that it is not true. It is a theorem of epistemic logic.

(4) can be substituted with variable of x. So x if and only if p–>Kp. (4) being necessarily true implies necessarily x. (4) and x are necessarily true.

(5) can be substituted with variable of y. So y if and only if Kp–>p. (5) being necessarily true implies necessarily y. (4) and y are necessarily true.

(6), based on substitutions of (4) & (5), takes on the form of x<–>y. Or we can say analytically true if and only if analytically true

x is analytic implies either necessarily x or necessarily not x. [](p–>Kp) v []~(p–>Kp).
y is analytic implies either necessarily y or necessarily not y. [](Kp–>p) v []~(Kp–>p).

(6) shows that we have collapsed any epistemic distinction between knowledge and truth. There is no difference between knowing something is true and something is true. What this specific proposition presents is that truth if and only if knowledge

Equivalency: ECpqCqp or ECNqNpCNpNq

(1) CpMp
(2) CMpp
(3) ECpMpCMpp

(4) CpKp
(5) CKpp
(6) ECpKpCKpp

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

 
%d bloggers like this: