RE: Why ontological arguments are illogical
August 9, 2012 at 10:33 pm
(This post was last modified: August 9, 2012 at 10:34 pm by genkaus.)
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: Yes, but the "perfect" amount of beauty might not be the most. The perfect amount of pizza isn't necessarily the largest amount of pizza.
But we are not talking about perfect amount of beauty, we are talking about perfect beauty.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: Yes, it does. Your turn.
No. it doesn't.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: Do you see my point? Yes, if you define "taking damage" to be a flaw, then flawlessness requires taking no damage. But why should "taking damage" be a flaw?
Because you take damage when the opponent's attack gets through your defense. A flawless defense does not allow that to happen. Ergo, taking damage = flaw.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: This is like saying that a "perfect" run of a game would be at max level. But lots of people try to beat the game at the lowest level possible. "Perfect" =/= "biggest".
Nonsense. The indicator is number of points garnered. You can get points at a higher rate at higher levels and you can end up with more points at lower level and getting further. So yes, a perfect game would take place at the higher level you get through.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: You're the one criticizing it, you tell me.
I'm telling you its not.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: Because if he didn't support his claim, I can reject it. Unsupported claims can be rejected.
He's not making a claim, he's making a statement - one common enough not to warrant independent support.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: No, that's my whole point; what you call "perfect" is somewhat arbitrary. It depends on what you're trying to 'optimize'.
No, what you call "perfect" is arbitrary, since I have actually laid out what it means and stayed consistent.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: Probably something like, "A quality is negative <=> the more one possesses it, the more unjustified suffering occurs"
Then no quality would be classified as positive or negative, because its effect is rarely consistent and depends much more on the context than the amount.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: Proof?
Mercy and justice.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: But "An entity can have all types of perfections" isn't an axiom. It was derived.
Really? Because it seems like Leibniz started with that premise.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: 1) How do you know that "Perfection can be analyzed" together with "An entity can have all types of perfections" entails a provable contradiction?
See the example of perfect morality and immorality.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: 2) How do you know that "Perfection cannot be analyzed" together with "An entity can have all types of perfections" entails an unprovable contradiction?
Starting analysis of perfection with the premise "An entity can have all types of perfections" entails a contradiction. Having the added premise of "Perfection cannot be analyzed" prevents any analysis and therefore proof, but the contradiction pursuant to the first would still entail.
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: Your argument is necessarily false.
Let A be your argument; then if A is true, "Perfection cannot be analyzed" together with "An entity can have all types of perfections" entails a contradiction, AND it cannot be proved that "Perfection cannot be analyzed" together with "An entity can have all types of perfections" entails a contradiction.
But if A is true, then it proves that "Perfection cannot be analyzed" together with "An entity can have all types of perfections" entails a contradiction. This is a contradiction. Therefore, A is false.
Do you see?
Let P = "Perfection can be analyzed", E = "An entity can have all types of perfections. Let F(x) = "X can be proved".
A states: "P and E => c (where c means 'contradiction'); ~P and E => ~F(~P and E => c); ~P and E => c." That is, if P and E are true, then there's a contradiction; if ~P and E are true, then there's a contradiction, but it can't be proved that there's a contradiction.
Suppose A true. Then under A, we know "~P and E => c", and we know ~F(~P and E => c). But then A is a proof that "~P and E => c". Hence F(~P and E => c). Thus A entails contradiction.
What are you talking about? What is "A"?
(August 9, 2012 at 9:58 pm)CliveStaples Wrote: Well, you didn't look at his actual argument. You looked at Stanford's one-sentence summation of his conclusion. So you might want to go look at his actual reasoning.
Is it any different from the one posted in OP?
(August 9, 2012 at 10:00 pm)RaphielDrake Wrote: Yeah, I can pretty much end this right here.
Prove perfection exists Clive, give me an example of something that is perfect that is demonstrable.
Anything. Then define what makes it perfect.
That's stupid. How is actual existence of relevant to an argument about god?