Another example involves the -]statement/-] “there exists”. When one -]instantiates /-]any statement in logic, one cannot instantiate any “there exists” statement afterwards with the same constant. [huh??]
Wrong:
(1) “There exists” is not even a statement in logic. It is merely an expression for the operator “Ex”
(2) Statements are not “instantiated,” they are merely uttered. Only properties are instantiated by objects, and those instantiations of properties by objects are what propositions and statements are
about.
Since I can instantiate anything with a “for all” statement, nothing stops me from instantiating “p” and “~p”. Syntax, for all his bluster (and some errors), **[which errors?] **actually gets it right. The very first statement I have made (there does not exist an x such that x) is self-contradictory. This, however, is simply one possible interpretation for the statement “nothing exists”. I prefer statements like -]"it is possible that nothing is real, or that nothing would be real/-].
This your problem. You are inventing meanings to suit your own purposes. But no one has to take you seriously. “Real” is just identical in meaning to “existent.”
For example, using (Rx) to mean “x is real”, you can say “Ex, Rx” (“there exists an x such that x is real”) and “~Ax, Rx” (“not all x’s are real”). Both of these are perfectly possible, without producing a self-contradiction.
First, you are treating “the real” as if it were a predicate. Why? Does it have a different meaning than “to exist”? You need to argue for this assumption. All respectable logicians will disagree with your proposed ambiguity between “real” and “existent.”
Second, “~(Ax) Rx” does not mean “not all x’s are real.” It means “it is not the case all x’s are real.” It is **logically equivalent **to saying “there exists an x such that x is not real.”
~(Ax) Rx <=> (Ex) ~Rx
So if “real” means “exist” then you are stating another contradiction.
The second statement would be processed “Ex, ~Rx”, and then instantiated “~Rp”.
What does “processed” mean in logic?
Once that one’s instantiated, another variable must be chosen for the first statement “Ex, Rx”, so we get “Rq”. Combining these two we get “Rq & ~Rp” which is not a contradiction.
First, “q” is not a variable; it’s a
constant. Only “x” “y” “z” are variables.
Second, if “real” means “existent” then this is a contradiction because your names “p” and “q” are presupposing the objects designated by those names exist. So how you can have a
non-real, but
existent thing is beyond me.
However, “Ax” is everything “for all x”, and so this would include all statements, including “p” and “~p”.
“Ax” can range over anything you want: objects, properties, propositions, sentences. But those things, whaterver they are, have to exist. For a statement using “Ax” to be true, false, or mean anything at all, it must range over really existent objects.