Hi!
I have made an exercise in sentential logic, and would like to hear, whether you think it is correctly done.
Task:
If God is willing to prevent evil, but unable to do so, he is impotent. If God is able to prevent evil but unwilling to do so, he is malevolent. Evil exist if and only if God is willing or unable to prevent it. God exists only if he is neither impotent nor malevolent. Therefore, if God exists then evil does not exist.
Solution: (perhaps)
The catch is that the argument is actually a tetralemma (not a trilemma, hvilke er en crucial premise). There are 2 important propositions
A = God is able to prevent evil, and W = God is willing to prevent evil,
I = God is impotent, M = God is malevolent, E = Evil exists, G = God exists
There are 2 truth values (true and false). So, there are 4 possible combinations and their resulting conclusions:
1. (A & ~W) → M
2. (W & ~A) → I
3. (A & W) → ~E
4. (~A & ~W) → ~G
This exhausts all possible worlds. Therefore the conclusion is
5. G → ~( M v I) therefore G → ~E.
The formalization of the above into one sentence (which I am uncertain of) are:
(A & ~W) → M, (W & ~A) → I, (A & W) → ~E, (~A & ~W) → ~G╞ G → ~E.
The Riddle of Epicurus
Re: The Riddle of Epicurus
Using math to prove thiesm?
Re: The Riddle of Epicurus
Yeah maths or logic will never dissuade anyone from blind faith. There's always a get out of any argument, hence religious apologetics. 
Edited to add: sorry UK.
Essentially there is only one premise in any argument of faith: do you believe or not, and if not why, and if so why. The Devil is in the details.
Further Edited to add:
The logical solution is to be agnostic, but then where does that leave you, if you are going to argue the philosophy.
Edited to add: sorry UK.
Essentially there is only one premise in any argument of faith: do you believe or not, and if not why, and if so why. The Devil is in the details.
Further Edited to add:
The logical solution is to be agnostic, but then where does that leave you, if you are going to argue the philosophy.
Last edited by Blaggard on Sat Jan 10, 2015 2:47 am, edited 5 times in total.
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Arising_uk
- Posts: 12259
- Joined: Wed Oct 17, 2007 2:31 am
Re: The Riddle of Epicurus
It's not Maths but Logic.
Re: The Riddle of Epicurus
The formalisation you provided at the end of your post is invalid. Firstly, here it is again:
(A & ~W) → M, (W & ~A) → I, (A & W) → ~E, (~A & ~W) → ~G╞ G → ~E.
Remember, that if an argument is valid, then it's conclusion must be true if its premises are true. Therefore, if it's possible to show that the premises of an argument are true and the conclusion false (without arriving at a contradiction), then the argument in question must be invalid.
So, let's assume that the conclusion of your formalisation is false: (G → ~E) = 0
For 'G → ~E' to be false, 'G' must be true and '~E' must be false, so: (G) = 1 and (~E) = 0
If (~E) = 0, then (E) = 1
We'll put these into a list:
1. (G) = 1
2. (E) = 1
Now, we have to try to make all of the premises true without contradicting (1) or (2) (so we can't encounter both (a)=0 and (a)=1 on this list, that would count as a contradiction). Let's start with '(A & ~W) → M', and then work through each premise one by one.
If ((A & ~W) → M) = 1, then either (A & ~W) = 0 or (M) = 1 (though not necessarily both). Let's assume that (M) = 1, this is sufficient to make '(A & ~W) → M' true:
3. (M) = 1
Next, we assume that ((W & ~A) → I) = 1. If ((W & ~A) → I) = 1, then either (W & ~A) = 0 or (I) = 1. Let's assume that (I) = 1:
4. (I) = 1.
Next, we assume that ((A & W) → ~E) = 1. So, either (A & W) = 0 or (~E) = 1. We don't want to make (~E) = 1, because that would contradict line (2) on the list, so instead we'll assume that (A & W) = 0. Now, if (A&W) = 0, then either (A) = 0 or (W) = 0. Let's assume that (W) = 0:
5. (W) = 0
Alright, we have one last premise: (~A & ~W) → ~G. Let's assume that ((~A & ~W) → ~G) = 1. So, either (~A & ~W) = 0 or (~G) = 1. We don't want to make (~G) = 1, because that would contradict line (1) on the list. Instead, we'll assume that (~A & ~W) = 0. Now, if (~A & ~W) = 0, then either (~A) = 0 or (~W) = 0. If (~W) = 0, then (W) = 1; but we can't do that, because (W) = 1 would contradict line (5) on the list. Instead, we'll assume that (~A) = 0. If (~A) = 0, then (A) = 1, so:
6. (A) = 1.
Now that we've finished making all of the premises true and the conclusion false, we can put the full list together:
1. (G) = 1
2. (E) = 1
3. (M) = 1
4. (I) = 1
5. (W) = 0
6. (A) = 1
This list constitutes a model in which the conclusion of your formalisation is false and, at the same time, all of the premises are true. Since this model does not contain any contradictions, your formalisation is invalid. However, when the premise 'G → ~(M v I)' is also taken into account, the argument is a valid one:
(A & ~W) → M, (W & ~A) → I, (A & W) → ~E, (~A & ~W) → ~G, G → ~(M v I)╞ G → ~E.
Aside from this, I have one other possible criticism of this particular formalisation of the problem of evil. Propositions 'G' and 'E' translate as "God exists" and "Evil exists" respectively. In both cases, some object or entity (in the former case it is "God", in the latter case it is "Evil") is predicated with "exists". That is to say, if taken literally, existence is treated in this formalisation as though it were a property that belongs to certain objects, rather than as though it were mere a quantifier. So, I'm slightly skeptical as to whether this argument can be adequately formalised using propositional logic; I think a first-order formalisation would be a much more suitable one.
(A & ~W) → M, (W & ~A) → I, (A & W) → ~E, (~A & ~W) → ~G╞ G → ~E.
Remember, that if an argument is valid, then it's conclusion must be true if its premises are true. Therefore, if it's possible to show that the premises of an argument are true and the conclusion false (without arriving at a contradiction), then the argument in question must be invalid.
So, let's assume that the conclusion of your formalisation is false: (G → ~E) = 0
For 'G → ~E' to be false, 'G' must be true and '~E' must be false, so: (G) = 1 and (~E) = 0
If (~E) = 0, then (E) = 1
We'll put these into a list:
1. (G) = 1
2. (E) = 1
Now, we have to try to make all of the premises true without contradicting (1) or (2) (so we can't encounter both (a)=0 and (a)=1 on this list, that would count as a contradiction). Let's start with '(A & ~W) → M', and then work through each premise one by one.
If ((A & ~W) → M) = 1, then either (A & ~W) = 0 or (M) = 1 (though not necessarily both). Let's assume that (M) = 1, this is sufficient to make '(A & ~W) → M' true:
3. (M) = 1
Next, we assume that ((W & ~A) → I) = 1. If ((W & ~A) → I) = 1, then either (W & ~A) = 0 or (I) = 1. Let's assume that (I) = 1:
4. (I) = 1.
Next, we assume that ((A & W) → ~E) = 1. So, either (A & W) = 0 or (~E) = 1. We don't want to make (~E) = 1, because that would contradict line (2) on the list, so instead we'll assume that (A & W) = 0. Now, if (A&W) = 0, then either (A) = 0 or (W) = 0. Let's assume that (W) = 0:
5. (W) = 0
Alright, we have one last premise: (~A & ~W) → ~G. Let's assume that ((~A & ~W) → ~G) = 1. So, either (~A & ~W) = 0 or (~G) = 1. We don't want to make (~G) = 1, because that would contradict line (1) on the list. Instead, we'll assume that (~A & ~W) = 0. Now, if (~A & ~W) = 0, then either (~A) = 0 or (~W) = 0. If (~W) = 0, then (W) = 1; but we can't do that, because (W) = 1 would contradict line (5) on the list. Instead, we'll assume that (~A) = 0. If (~A) = 0, then (A) = 1, so:
6. (A) = 1.
Now that we've finished making all of the premises true and the conclusion false, we can put the full list together:
1. (G) = 1
2. (E) = 1
3. (M) = 1
4. (I) = 1
5. (W) = 0
6. (A) = 1
This list constitutes a model in which the conclusion of your formalisation is false and, at the same time, all of the premises are true. Since this model does not contain any contradictions, your formalisation is invalid. However, when the premise 'G → ~(M v I)' is also taken into account, the argument is a valid one:
(A & ~W) → M, (W & ~A) → I, (A & W) → ~E, (~A & ~W) → ~G, G → ~(M v I)╞ G → ~E.
Aside from this, I have one other possible criticism of this particular formalisation of the problem of evil. Propositions 'G' and 'E' translate as "God exists" and "Evil exists" respectively. In both cases, some object or entity (in the former case it is "God", in the latter case it is "Evil") is predicated with "exists". That is to say, if taken literally, existence is treated in this formalisation as though it were a property that belongs to certain objects, rather than as though it were mere a quantifier. So, I'm slightly skeptical as to whether this argument can be adequately formalised using propositional logic; I think a first-order formalisation would be a much more suitable one.
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GreatandWiseTrixie
- Posts: 1543
- Joined: Tue Feb 03, 2015 9:51 pm
Re: The Riddle of Epicurus
I'm inclined to agree, math (or math style logic) is not the best solution for this little thing of yours.sjeff70 wrote:Using math to prove thiesm?
And, your argument lacks a solid foundation. What is good and evil to a God? Those are human terms. The statements are dead upon arrival. I'd say they do not prove a malevolent God, but a neutral one.