Re: Prove axiom in S5
From: muxol (-knowledge-_at_excite-dot-com.no-spam.invalid)
Date: 03/01/05
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Date: 1 Mar 2005 05:34:26 -0600
> William Elliotwrote:
On Mon, 28 Feb 2005, muxol wrote:
>
> A.S.wrote:
> Does anybody know how to prove the following using Natural
Deduction
> in S5 (using [] for the box, and <> for the diamond)?
>
> ([]A -> A), (A -> []<>A), ([]A -> [][]A) [single
> turnstile, S5] (<>A
> -> []<>A)<>A)<>A)<>A)[/quote:6d1a1426b3]
1. A -> []<>A (axiom)
2. <>A -> []<><>A (sub instance of 1 with
A as the A)
3. <><>A -> <>A (alternative axiom of []A -
[][]A -- see end note)
4. <><>A -> []<><>A (2,3 by
truth-functions)
5. <>A -> []<>A (4 with the A as the
<>A)[/quote:34299f8a3b]
Baloney. You can't do that substitution.
That's a logical as saying
0 <= x^2, thus 0 < x by replacing x with x^2.
> Note:
>
> 1. []A ->[][]A
> 2. ~[][]A -> ~[]A (1 by truth-functions (contrapositive))
> 3. <>~[]A -> <>~A (2)
> 4. <><>~A -> <>~A (3)
> 5. <><>A -> <>A (4)
>
> QED.
>
>
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> [/quote:34299f8a3b]
Do you mean the sub in line 5? I agree. But the sub in line 2 is
perfectly fine.
However, reiterations of modal operators are redundant -- i.e.
<><>A iff <>A, [][]A iff []A, and hence
<><>A -> []<><>A iff <>A ->
[]<>A. 4 to 5 is legitimate in S5.
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