- #1

- 92

- 0

(~) = not, (v) = or, (&) = and, (->) = implies (<->) = equivalence,

[] = necessarily, <> = possibly, (=>) = strict implication,

(p => q) =df [](p -> q)

<>p =df ~[]~p.

g = god exists.

The argument is thus:

1. g => []g (premise)

2. []g v ~[]g

3. ~[]g => []~[]g

4. []g v []~[]g

5. []~[]g => []~g

6. []g v []~g

7. <>g (premise)

8. []g

9. []g => g

10. g

This argument is valid but not sound.

It proves that: ((g => []g) & <>g) -> g, is necessarily true, ..nothing else.

The argument is true for any proposition p.

1. g => []g

7. <>g

:.

10. g

A. (g => []g) <-> [](g -> []g)

B. [](g -> []g) <-> (<>g -> []g),

C. (g => []g) <-> (<>g -> []g)

Note: A, B, C, are theorems of modal logic (S5).

Because of C, the argument becomes:

1. <>g -> []g

7. <>g

:.8. []g

8. []g

9. []g -> g

:.10. g

If we substitute ~g for g, we get the atheists' side of it.

1a. <>~g -> []~g

7a. <>~g

:. 8a. []~g

8a. []~g

9a. []~g -> ~g

:.

10a. ~g.

This argument has two other equivalent variations.

1. [](g -> []g) & <>g .-> g

2. [](<>g -> g) & <>g .-> g

3. (<>g -> []g) & <>g .-> g

Once we realise that: [](p -> []p) <-> (<>p -> []p),

and [](<>p -> p) <-> (<>p -> []p), we can see that each

argument is equivalent to 3.

Hartshorne was wrong to assert that this argument proves that g (god exists) is true.

It seems that Theists need only show that 'God does exists' is possible

in order to prove that it is necessary or that it is true.

And, that Atheists need only to show that 'God does not exists' is possible

in order to prove that it is necessary or that it is true.

Note: <>(god exists) & <>(god does not exist), is contradictory.

What do you think?

Owen