Saturday, June 15, 2013

Predicate logic wholesale

Passes 4-0. — Quirck

Adminned at 16 Jun 2013 19:43:46 UTC

Add the following Axioms:

* [0 && 0] = 0
* [0 && 1] = 0
* [1 && 0] = 0
* [1 && 1] = 1
* [0 || 0] = 0
* [0 || 1] = 1
* [1 || 0] = 1
* [1 || 1] = 1
* [0 -> 0] = 1
* [0 -> 1] = 1
* [1 -> 0] = 0
* [1 -> 1] = 1
* [~ 0] = 1
* [~ 1] = 0

If quantifiers are added, perhaps we can add axioms describing them,

Comments

Sphinx:

15-06-2013 20:37:00 UTC

I don’t know, I personally would prefer !x as NOT, but else this seems good to me. for

Sphinx:

15-06-2013 20:38:41 UTC

ACtually, I don’t think this proposal would do anything, would it?

Sphinx:

15-06-2013 20:40:48 UTC

Right, I’m tired and should go to bed, ignore my last comment.

quirck: he/him

15-06-2013 20:43:14 UTC

“Any Atom can add an item to either list by submitting an entry in the “Proof” category that states the Formula to be added, which section to add it to <...>”
Does it mean that these axioms should already be listed on the Truths page? The addition of the axioms doesn’t require the Proof to pass.. And enacting the Proof does not imply any actions other than reporting the final tally of votes…

Sphinx:

15-06-2013 20:57:06 UTC

That looks like something that should be fixed…

Tavros:

15-06-2013 21:08:03 UTC

I think the intent of the rule is clear, and I’m fine with interpreting it as intended,

kikar:

16-06-2013 01:53:45 UTC

for I wish there were a more compact way to write this, but I guess there isn’t?

Skju:

16-06-2013 03:42:38 UTC

Actually…
~ p = p - p
p & q = [p - p] - [q - q]
p || q = ~ [p - q]
p -> q = [~ p] || q

And we sure should patch that Proof thing.

quirck: he/him

16-06-2013 06:10:20 UTC

for