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# Thread: The Ontological Argument, part I: Necessity and Possibility

1. ## The Ontological Argument, part I: Necessity and Possibility

NOTE: Though I use the word "Ontological argument", I am not referring to the idiotic one made by St. Anselm. His argument is terrible. This argument is much more interesting. A similar argument was put forth by Kurt Godel (Best known for his "Incompleteness theorem" and he was also one of Einstein's best friends), and more recently by Alvin Plantinga. I'll play Devil's Advocate.

Part I: Necessity and Possibility

First, I need to discuss some basic facts about the structure and grammar of logic, which shall comprise "Part I: Necessity and Possibility" (Part II: Maximal Excellence and the argument will come after we can agree upon these premises). The basis of this argument rests on a very counter-intuitive fact of logic, arising from a discipline known as modal logic. Unfortunately, in order to understand this principle, you need to have a basic understanding of symbolic logic. So, let's begin:

Firstly, you should reflect on the fact that if a proposition is not, not true, then it is true. This follows naturally from the idea that a proposition can either be true or false; therefore, if it is not false, then it must be true. Let's call a given proposition "P", and define "P is false" or "Not P" under the symbolic notation "~P". Therefore, our given principle described above is ("=" means "is logically equivalent to"):

Axiom1: ~(~P) = P

Now we begin modal logic. Modal logic deals with propositions about "necessity" and "possibility". For example, if both "If A, then B" and "A is true" are true statements, then it follows necessarily that B is also true (i.e., "B is necessarily true"). Possibility is fairly intuitive; we understand when things are possible or impossible. Symbolically, let's define "M(P)" to mean "Possibly P" (or, "P is possibly true") and "L(P)" to mean "Necessarily P" (or, again, "P is necessarily true").

Secondly, we can reflect on the fact that there is an important relationship between the "necessity" (L) and "possibility" (M). The relationship works like this: "A proposition is necessarily true, if and only if it is not possibly not true." Which corresponds directly to what we believe about necessity --a necessary proposition cannot possibly be false. Or, symbolically:

Axiom2: L(P) = ~M(~P)

And the same analogous statement is true for possibility:

Axiom3: M(P) = ~L(~P)

Or, in English, "A statement is possible, if and only if it is not necessarily false." Or even more simply: "If something isn't definitively false, then it is still open to being true." and "If something isn't possibly any other way, then it must be true."

Take a little while to reflect on those. There's actually a larger logical theory being used here (Known as De Morgan duals, after Augustus De Morgan) but I don't really need to get into that theory in order to proceed. Okay, do you follow this so far? If not, then I suggest you re-read. If yes, then continue along.

So, I only have one more axiomatic proposition for you to consider and reflect on:

"Something is possibly true, if and only if it is necessarily possibly true." EDIT: I give a full proof of this in my second post on this thread.

Which is to say, if something is necessarily possibly true, then obviously it is possibly true. So this statement above is true. And in symbolic logic:

Axiom4: L(M(P)) = M(P)

So, taken all together, let's evaluate a whole different statement.

Theorem 5: M(L(P)) = ~L(~(~M~(P))) = ~L(M(~P)) = ~M(~P) = L(P)

(Convert M into ~L~ on the left, and L into ~M~ on the right. These are true by definition. Then, ~(~Q) = Q, so we get rid of the two ~ in between L and M. Then, LM(Q) = M(Q), so you can get rid of the L. Then this is the same thing as the definition of L(P). QED.)

So, more compactly:

Theorem 5: M(L(P)) = L(P)

In English, "If and only if it is possibly true that P is necessary, then P is necessary." (And obviously, if P is necessary, then P is true)

Are we all on the same page here, ladies and gentlemen? Yes, this is an annoying and strange fact. But it is a fact, I promise.

(By-the-by, this is not the incorrect part of the debate, this is all absolutely, certainly true. There's simply the question of it being incredibly pathological and counter-intuitive, so you will need a while to reflect upon this odd nature of logic)  Reply With Quote

2. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by GoldPhoenix [So, taken all together, let's evaluate a whole different statement.

Theorem 5: M(L(P)) = ~L(~(~M~(P))) = ~L(M(~P)) = ~M(~P) = L(P)

(Convert M into ~L~ on the left, and L into ~M~ on the right. These are true by definition. Then, ~(~Q) = Q, so we get rid of the two ~ in between L and M. Then, LM(Q) = M(Q), so you can get rid of the L. Then this is the same thing as the definition of L(P). QED.)
The bolded portions appear to be in error. While you have established that L(M(P)) = M(P), you have not established that ~L(M(~P)) = ~M(~P). The latter does not follow from the former: The former means "If it is necessary that P is possibly true, then P is possibly true". But the latter means "If it is not necessary that P is possibly false, then it is impossible that P is false", which is a different thing altogether.

Or to use symbolic language, you cannot just take out the "L" when it is being modified by "~", and tack on the "~" onto "M", for this implies that something that is not necessarily true is also not possibly true (i.e. it is necessarily false), which you have not proven (and which, in fact, is false).

To illustrate this, let's represent (M(~P)) as Q. According to you,
~L(Q) = ~Q.
In English, this means that "If Q is not necessarily true, then Q is false." Clearly, this does not logically follow - even if Q is not necessarily true, it might still be possibly true. Therefore,
~L(Q) =/= ~Q
And since Q = (M(~P)),
~L(M(~P)) =/= ~M(~P)
Thus disproving the bolded portion of your equation.  Reply With Quote

3. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by Trendem The bolded portions appear to be in error. While you have established that L(M(P)) = M(P), you have not established that ~L(M(~P)) = ~M(~P). The latter does not follow from the former: The former means "If it is necessary that P is possibly true, then P is possibly true". But the latter means "If it is not necessary that P is possibly false, then it is impossible that P is false", which is a different thing altogether.
If X=Y
~X=~Y

L(M(P)) = M(P)
~[L(M(P))] = ~[M(P)]

Do you agree that M(P) = L(M(P))? That is, "If P is possibly true, then it is not possibly not true that P is possibly true."

Do you agree that ~M(P) = ~L(M(P))? That is, "P is not possibly true" = "It is not necessarily true that P is possibly true".

Do you agree that M(~P) = L(M(~P))? That is, "If P is possibly not true, then it is not possibly not true that P is possibly false."

Do you agree that ~M(~P) = ~L(M(~P))?

Or: "P is necessarily true" = "It is not necessarily true that P is possibly false"

~L(M(~P)) = ~(~M~(M(~P))) [using L = ~M~] = M~(M(~P)) = M(~M(~P))

If P is necessarily true, then it is possible that P is necessarily true.

Or to use symbolic language, you cannot just take out the "L" when it is being modified by "~", and tack on the "~" onto "M", for this implies that something that is not necessarily true is also not possibly true (i.e. it is necessarily false), which you have not proven (and which, in fact, is false).

First, he's not replacing "L" with "~M", he's replacing it with "~M~". Thus:

L(P) = ~M~(P)

~L(P) != ~M(P) (This is what you claimed he said)

~L(P) = ~(~M~(P)) = M(~P) (This is what is in fact true)

Or: "If something is not necessarily true, it is possibly not true."

Something that is not necessarily true is possibly not true. You replace "L[x]" with "~M~[x]", not "~M[x]".

To illustrate this, let's represent (M(~P)) as Q. According to you,
~L(Q) = ~Q.
In English, this means that "If Q is not necessarily true, then Q is false." Clearly, this does not logically follow - even if Q is not necessarily true, it might still be possibly true. Therefore,
~L(Q) =/= ~Q
And since Q = (M(~P)),
~L(M(~P)) =/= ~M(~P)
Thus disproving the bolded portion of your equation.
To use your argument elsewhere:

If you accept M(P) = LM(P)

Substitute Q= M(P)

Q=LQ

If Q is true, Q is necessarily true.

Your "substitution" method is not truth-preserving.

Or rather, statements about Q can only be evaluated by applying them to what Q has been substituted for. A statement might be false of Q as a proposition, but not of the proposition Q was substituted for.

Which is to say:

If Q=M(P),
Q = LQ

If Q = P
Q != LQ  Reply With Quote

4. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by Autolykos Here is the corrected substitution:

M(L(P)) = ~L(~L(P)) = ~L(~(~M(~P))) = ~L(M(~P)) = M(P)

Of course, this reduces to:

M(L(P)) = M(P)

Using verbal symbolism (i.e. plain English), this equates to "if it is possible that it is necessary that P is true, then it is possible that P is true".

To help understand the final step in the substitution, ~L(M(~P)) = M(P), let's first find out what the negation of L(P) is equal to in terms of M:

~(L(P)) = ~(~M(~P)) (negation of your Axiom 2)
~L(P) = M(~P) (distribution of negation)

In other words, if it is not necessary that P is true, then it is possible that P is false.

Now to make things simpler, let's use a new symbol I (for "inverse") defined as I = ~P:

~L(M(I)) = M(~I) (given the above)

Substituting ~P for I, we now get:

~L(M(~P)) = M(P)
EDIT: Forgot to add an additional factor of "~", so my conclusion was erroneous.

Whose substitution you're correcting? GP's substitutions were fine.  Reply With Quote

5. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by CliveStaples ~L(M(~P)) = ~M~(M(~P) = ~M(L(P))

You're saying:

~ML(P) = M(P)

This is not true.

"If P is possibly true, it is not possible that P is necessarily true." This is obviously false; if P is necessarily true, then P is possibly true by definition.

Or:

"If it is not possible that P is necessarily true, P is possibly true."

This is not true. If P is not possibly necessarily true, P could be necessarily false; you cannot say that P is possibly true simply because ~ML(P).

And I'm not sure whose substitution you're correcting. GP's substitutions were fine.
I looked over my work again and saw that it had a fatal error. Basically I jumped the gun on working through the substitutions. If ~L(P) = M(~P), then ~L(M(P)) =/= M(~P). This is what I get for trying to multitask. Let me go on the record and say that I stand corrected. The substitutions that GP presented are definitely valid.  Reply With Quote

6. ## Re: The Ontological Argument, part I: Necessity and Possibility

On another note, I think the reason why that final proposition (namely M(L(P)) = L(P)) seems counter-intuitive is because people intuitively think that, if something can be true, it can also be false. In other words, the truth value of the something is not currently known (i.e. at the time of thought/speech).  Reply With Quote

7. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by Autolykos I looked over my work again and saw that it had a fatal error. Basically I jumped the gun on working through the substitutions. If ~L(P) = M(~P), then ~L(M(P)) =/= M(~P). This is what I get for trying to multitask. Let me go on the record and say that I stand corrected. The substitutions that GP presented are definitely valid.
Funnily enough, I did the same thing.

You wrote:

~L(M(~P)) = M(P)

And I translated it as:

~M(~(M(~P)) = M(P)

When in fact it is:

~(~M(~M(~P))) = M(P)

~L(M(P)) = M(~P)
~~M~M(P)
M~MP
ML~P

If P is possibly necessarily false, P is possibly false.
If P is possibly false, P is possibly necessarily false. (If P is possibly false, P is not necessarily possibly true)

This seems true by definition, doesn't it?

In any case, back to the ontological argument...  Reply With Quote

8. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by Autolykos On another note, I think the reason why that final proposition (namely M(L(P)) = L(P)) seems counter-intuitive is because people intuitively think that, if something can be true, it can also be false. In other words, the truth value of the something is not currently known (i.e. at the time of thought/speech).
This is true. (That people intuitively think that it is true, I mean.)

Most people think that:

M(P) <=> M(~P)

But this is not in fact the case.  Reply With Quote

9. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by CliveStaples This is true. (That people intuitively think that it is true, I mean.)

Most people think that:

M(P) <=> M(~P)

But this is not in fact the case.
This may sound like a dumb question, but why not? Perhaps we need a higher-order logic to capture the meaning of language constructs like the English modal can?  Reply With Quote

10. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by Autolykos This may sound like a dumb question, but why not? Perhaps we need a higher-order logic to capture the meaning of language constructs like the English modal can?
The modal demonstration goes like this:

L(P) => M(P) "If something is necessarily true, it must also be possibly true."

If M(P) = M(~P), then:

L(P) => M(~P)

But:

L(P) = ~(M(~P))

Therefore:

M(P) != M(~P)

When we say "P is possibly true", we aren't saying "P is possibly false". We're saying "P is not necessarily false". If by saying "P is possibly true", we implied "P is possibly false", we would in fact be asserting "P is neither necessarily true nor necessarily false." But "P is possibly true" still allows for "P is necessarily true".  Reply With Quote

11. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by Trendem The bolded portions appear to be in error. While you have established that L(M(P)) = M(P), you have not established that ~L(M(~P)) = ~M(~P). The latter does not follow from the former: The former means "If it is necessary that P is possibly true, then P is possibly true". But the latter means "If it is not necessary that P is possibly false, then it is impossible that P is false", which is a different thing altogether.
If I'm understanding you correctly in this critique, your problem lies in the idea that I didn't say "If and only if A, then B" and I accidentally only said "If A, then B". (If it weren't "if and only if", this particular statement you made above would be true, and you would be right that my entire argument would be shot) To illustrate for those who don't follow: "If and only if A, then B" (logical equivalency) is the same thing as saying "If A, then B" AND "If B, then A." So to prove my argument both ways is to prove that the claims are logically equivalent (which makes my entire argument true):

It is simply enough to prove by contradiction that the converse (B=>A) you're holding question, M(P) => L(M(P)) must be true.

So, as I said, proof by contradiction. Assume that:

~(M(P) => L(M(P))) <=> (M(P) /\ ~L(M(P))

Because this is the definition of the negated implication. (NOTE: "A /\ B" means "A and B are true") So,

(M(P) ^ ~L(M(P)) <=> ~L(~P) ^ M(L(~P)

By just turning the original M into ~L~, and the original L into ~M~, canceling out the ~~. So in order to say that M(P) =/=> L(M(P)), you'd have to say:

"~L(~P) /\ M(L(~P)", which if we define "A <=> L(~P)", then we get:

"~A /\ M(A)", or in English "Not A and possibly A"

This is, obviously, absurd and quite obviously false for any proposition A. Something cannot both be possibly true AND be false at the same time. Ergo, we have:

M(P) => L(M(P))

And trivially for the non-converse, (i.e. A=>B ),

L(M(P)) => M(P)

(Obviously if something is necessarily possibly true, then it's possibly true)

Therefore, "M(P) = L(M(P))". QED. Originally Posted by Trendem
Or to use symbolic language, you cannot just take out the "L" when it is being modified by "~", and tack on the "~" onto "M", for this implies that something that is not necessarily true is also not possibly true (i.e. it is necessarily false), which you have not proven (and which, in fact, is false).
Actually, as I've shown, it is demonstrably true, not false. Going back to what you said before:

~L(M(~P)) = ~M(~P)

Is obviously true. If L(M(~P) = M(~P), then, clearly, ~(L(M(~P)) = ~(M(~P)), by the negation relation:

"A = B" <=> "~A = ~B", and this is trivially true. Originally Posted by Tren
To illustrate this, let's represent (M(~P)) as Q. According to you,
~L(Q) = ~Q.
In English, this means that "If Q is not necessarily true, then Q is false." Clearly, this does not logically follow - even if Q is not necessarily true, it might still be possibly true. Therefore,
~L(Q) =/= ~Q
And since Q = (M(~P)),
~L(M(~P)) =/= ~M(~P)
Thus disproving the bolded portion of your equation.
Unfortunately, this is predicated upon a sweeping generalization fallacy, and, consequently, a strawman fallacy. Fully demonstrated:

Def. (M(~P)) = Q.
Thm1. ~L(Q) = ~Q.

But, take note of the fact that Theorem 1 is true if and only if this particular Q is defined as a modal possibility statement (e.g. Q = M(P).). And the equivalency will hold, too, so long as Q is such a statement. But you immediately jumped the gun and said "Well, if Q is any statement" (I never said this, hence the strawman, and when strawmaning me you obviously committed a hasty generalization fallacy by equating all forms of statements with a specific modal form of statements) in the very next line: Originally Posted by Tren
In English, this means that "If Q is not necessarily true, then Q is false." Clearly, this does not logically follow - even if Q is not necessarily true, it might still be possibly true. Therefore,
~L(Q) =/= ~Q
And since Q = (M(~P)),
~L(M(~P)) =/= ~M(~P)[/I]
But I never said that, Trendem. I said that only if Q is a modal possibility statement then the equality "~L(Q) => ~Q" is true. And that equality is true, for the given Q, but not any others. But you're trying to say "Oh, but for other kinds of propositions, this isn't true." Great, but I'm not talking about those kinds of propositions. So that was a completely erroneous counter-proof.  Reply With Quote

12. ## Re: The Ontological Argument, part I: Necessity and Possibility

Caveat: I am new to modal logic (I just looked it up when GP posted the thread), so please bear with me if I exhibit unfamiliarity with certain terms and concepts. Originally Posted by CliveStaples Do you agree that M(P) = L(M(P))? That is, "If P is possibly true, then it is not possibly not true that P is possibly true."

Do you agree that ~M(P) = ~L(M(P))? That is, "P is not possibly true" = "It is not necessarily true that P is possibly true".

Do you agree that M(~P) = L(M(~P))? That is, "If P is possibly not true, then it is not possibly not true that P is possibly false."

Do you agree that ~M(~P) = ~L(M(~P))?

Or: "P is necessarily true" = "It is not necessarily true that P is possibly false"
Yes to all. Originally Posted by CliveStaples First, he's not replacing "L" with "~M", he's replacing it with "~M~". Thus:

L(P) = ~M~(P)

~L(P) != ~M(P) (This is what you claimed he said)

~L(P) = ~(~M~(P)) = M(~P) (This is what is in fact true)
But that is not what GP wrote. He wrote:
~L(M(~P)) = ~M(~P)
I do not see where he substituted "L" with "~M~". I think GP's explanation makes more sense - if L(M(P)) is a necessary condition of M(P), then ~L(M(P)) = ~M(P). Originally Posted by CliveStaples Or rather, statements about Q can only be evaluated by applying them to what Q has been substituted for. A statement might be false of Q as a proposition, but not of the proposition Q was substituted for.
Conceded.  Reply With Quote

13. ## Re: The Ontological Argument, part I: Necessity and Possibility Originally Posted by GoldPhoenix If I'm understanding you correctly in this critique, your problem lies in the idea that I didn't say "If and only if A, then B" and I accidentally only said "If A, then B". (If it weren't "if and only if", this particular statement you made above would be true, and you would be right that my entire argument would be shot) To illustrate for those who don't follow: "If and only if A, then B" (logical equivalency) is the same thing as saying "If A, then B" AND "If B, then A." So to prove my argument both ways is to prove that the claims are logically equivalent (which makes my entire argument true):

It is simply enough to prove by contradiction that the converse (B=>A) you're holding question, M(P) => L(M(P)) must be true.

So, as I said, proof by contradiction. Assume that:

~(M(P) => L(M(P))) <=> (M(P) /\ ~L(M(P))

Because this is the definition of the negated implication. (NOTE: "A /\ B" means "A and B are true") So,

(M(P) ^ ~L(M(P)) <=> ~L(~P) ^ M(L(~P)

By just turning the original M into ~L~, and the original L into ~M~, canceling out the ~~. So in order to say that M(P) =/=> L(M(P)), you'd have to say:

"~L(~P) /\ M(L(~P)", which if we define "A <=> L(~P)", then we get:

"~A /\ M(A)", or in English "Not A and possibly A"

This is, obviously, absurd and quite obviously false for any proposition A. Something cannot both be possibly true AND be false at the same time. Ergo, we have:

M(P) => L(M(P))

And trivially for the non-converse, (i.e. A=>B ),

L(M(P)) => M(P)

(Obviously if something is necessarily possibly true, then it's possibly true)

Therefore, "M(P) = L(M(P))". QED.

Actually, as I've shown, it is demonstrably true, not false. Going back to what you said before:

~L(M(~P)) = ~M(~P)

Is obviously true. If L(M(~P) = M(~P), then, clearly, ~(L(M(~P)) = ~(M(~P)), by the negation relation:

"A = B" <=> "~A = ~B", and this is trivially true.

Unfortunately, this is predicated upon a sweeping generalization fallacy, and, consequently, a strawman fallacy. Fully demonstrated:

Def. (M(~P)) = Q.
Thm1. ~L(Q) = ~Q.

But, take note of the fact that Theorem 1 is true if and only if this particular Q is defined as a modal possibility statement (e.g. Q = M(P).). And the equivalency will hold, too, so long as Q is such a statement. But you immediately jumped the gun and said "Well, if Q is any statement" (I never said this, hence the strawman, and when strawmaning me you obviously committed a hasty generalization fallacy by equating all forms of statements with a specific modal form of statements) in the very next line:

But I never said that, Trendem. I said that only if Q is a modal possibility statement then the equality "~L(Q) => ~Q" is true. And that equality is true, for the given Q, but not any others. But you're trying to say "Oh, but for other kinds of propositions, this isn't true." Great, but I'm not talking about those kinds of propositions. So that was a completely erroneous counter-proof.
Thanks, I get it now. Argument conceded; you can now go on to the Ontological Argument.   Reply With Quote

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