I know that they are equivalent statements from truth tables and common sense, but I have to come to this conclusion via only six mechanisms: elimination of an and; elimination of an or; introduction of an and; introduction of an or; introduction of a contradiction; elimination of a contradiction. To me, this result is so "duh!" but I can't figure out a valid proof. I can't use anything like de Morgan's laws as ... that would be beyond the mechanisms given me. John Riemann Soong (talk) 00:43, 28 September 2009 (UTC)[reply]

Does this help? I don't really understand your allowed operations but it looks to me like you're trying to prove a restricted statement of De Morgan's Laws of which the former link looks to provide a valid treatment. (Technically it shows the other part of De Mogan's Laws but it does suggest a way for you to proceed.) Martlet1215 (talk) 01:15, 28 September 2009 (UTC)[reply]

I think he means prove ¬(A ∨ B) ⇒ ¬A ∧ ¬B. ~~ Dr Dec (Talk) ~~ 11:13, 28 September 2009 (UTC)[reply]

(1−A)(1−B) = 1−(A+B−AB) is an algebraic identity equivalent to your logical identity, as ~A = 1−A and A/\B = AB and A\/B = A+B−AB. Bo Jacoby (talk) 06:21, 28 September 2009 (UTC).[reply]

Could you explain in more detail these "mechanisms" you are permitted to use? It's unclear what proof technique you are using. Is it a form of natural deduction? If so, what are "introduction of a contradiction" and "elimination of a contradiction"? Maelin (Talk | Contribs) 11:41, 28 September 2009 (UTC)[reply]

(e/c) It's not entirely clear what proof system are you using, but the names of the rules suggest that it's a kind of natural deduction. Nevertheless I'm not quite sure what exactly do you mean by "introduction/elimination of a contradiction". Here's a proof (in sequential form, for typographical reasons):

1. ¬(A ∨ B)       [assumption]
2. A              [temporary assumption]
3. A ∨ B          [∨-introduction from 2]
4. ¬A             [reductio ad absurdum from 1 and 3, discarding assumption 2]
5. B              [temporary assumption]
6. A ∨ B          [∨-introduction from 5]
7. ¬B             [reductio ad absurdum from 1 and 6, discarding assumption 5]
8. ¬A ∧ ¬B        [∧-introduction from 4 and 7]

If your rules for negation are different, it shouldn't be hard to adjust it; this form of De Morgan's law is intuitionistically valid, hence it should be provable using any reasonable version of negation rules. — Emil J. 11:48, 28 September 2009 (UTC)[reply]

I remember reading an article where a person would sell a $1 coin at a market/auction/something and people would "buy" the $1 coin. It shows how a person is forced to outbid the other bidder and thus eventually ends up paying more than the amount of the coin. Know what I'm talking about? :) Deon555 (talk) 06:49, 28 September 2009 (UTC)[reply]

Yep - and a lucky guess led me to Dollar auction. AndrewWTaylor (talk) 07:41, 28 September 2009 (UTC)[reply]

Haha, thanks Andrew - I kept searching dollar coin (instead of dollar 'note') and thus returning nothing :). Thanks for your help. Deon555 (talk) 10:54, 28 September 2009 (UTC)[reply]

Hi, I have a question about conformal maps from the surface of a sphere unto a square.

Is there a unique mapping (other than for choice of centre and orientation) that can achieve this? Or are there many such mappings?

Thanks.--Leon (talk) 21:29, 28 September 2009 (UTC)[reply]

See conformal map projections. ~~ Dr Dec (Talk) ~~ 21:40, 28 September 2009 (UTC)[reply]

I fail to see how this helps. This, this and this are all essentially the same, and none of the other conformal projections given map to squares.--Leon (talk) 21:44, 28 September 2009 (UTC)[reply]

Try following the links within the articles. I found this link within a link quite useful. If you want a better answer then perhaps you should pose a better question. For example, what does a map of A unto B mean? You say that the Peirce quincuncial projection, the Adams hemisphere-in-a-square_projection and the Guyou hemisphere-in-a-square projection are "essentially the same", well, what is your equivalence relation between maps? Your question was rather vague. ~~ Dr Dec (Talk) ~~ 22:27, 28 September 2009 (UTC)[reply]

The question seems pretty well defined. He wants a conformal bijection for the sphere to the square and two maps f and g are equivalent if f = gσ where σ is some rotation or improper rotation of the sphere. I don't know the answer by the way. Rckrone (talk) 03:44, 29 September 2009 (UTC)[reply]

The Peirce quincuncial projection is conformal except at four points - at the mid-point of each side of the square, angles are doubled when you go from the square to the sphere. You can see this if you look at the image of the equator, which makes a 90 degree turn at the mid-point of each side of the square. I think that is the most nearly conformal mapping that is possible between a square and a sphere. Gandalf61 (talk) 09:19, 29 September 2009 (UTC)[reply]