User talk:WillemienH

Wallace and Gromit: The Curse of the Were-Rabbit
Hi -- I removed Category:Fictional hares and rabbits from Wallace & Gromit: The Curse of the Were-Rabbit because the article itself is not about a fictional hare or rabbit. Trivialist (talk) 03:47, 11 January 2015 (UTC)

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Re: Lobachevsky Formula Redirect
Hello. I saw your message on my talk page. I was unsure if you would see if I replied there, so I'm leaving you a message here on your talk page. In any case, you did a good job with your edits. The only small issue was your initial edit to Angle of Parallelism was just a link to the intended target, not actually a redirect. To create a redirect, make this statement the first thing on a page.
 * #redirect Target of redirect

Otherwise, like I said, good job. Don't worry about being a beginner - we were all new once. :) – Majora4  (leave a message)   20:09, 8 February 2015 (UTC)

Teahouse
Hello! At the Teahouse, new questions go at the top. It is a special page. Just a friendly pointer. -DangerousJXD (talk) 05:11, 3 April 2015 (UTC)

Reference errors on 12 April
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Hyperbolic functions second derivatives
You made a minor omission in this edit. ;) CiaPan (talk) 21:07, 15 April 2015 (UTC)

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Genocides in history
Please see Talk:Genocides in history

Also if you post to a talk page and the posting includes footnotes please place a reflist template before your signature, so that your footnotes are kept in the same place as your posting.

-- PBS (talk) 09:42, 11 May 2015 (UTC)

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Spelling the right word
You walk through a door. Pigs eat out of a trough. JRSpriggs (talk) 15:35, 1 August 2015 (UTC)

Your GA nomination of Hyperbolic geometry
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Your GA nomination of Hyperbolic geometry
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Possible error in category linking
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ArbCom elections are now open!
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Hyperbolic triangles for Lorentz transformations (non-collinear boosts)
Hi WillemienH. I notice you are actively editing articles on hyperbolic geometry and hyperbolic triangles. I have been extensively rewriting Lorentz transformation (in fact the current version is almost entirely YohanN7's and my work). If you are interested and have the time, would you like to contribute on the connection between hyperbolic triangles and rapidity for two boosts in different directions? No obligations, just a thought.

You may already know this, but here is the background. For two boosts in the same direction, the rapidities add to get the rapidity of the overall boost. For two boosts in different directions, the Lorentz transformation is not a single boost, but a boost followed/preceded by a rotation.

For ease of visualization take 2 spatial dimensions and 1 time dimension for the spacetime diagram. The rapidity of a boost traces out a curve [in a hyperbolic sheet of constant time but varying spatial coordinates, a hyperbola (ct)2 - x2 - y2 = (ct&prime;)2] from one time axis to a second boosted time axis, and a third frame boosted relative to the second traces another curve in the surface. These two boosts form two edges of a hyperbolic triangle, the third edge is the resultant rapidity (but since the angles of a hyperbolic triangle are different to a flat Euclidean one, the rotation must be included somehow). I think this is correct, still looking for some good sources on this, a start has been made at Talk:Lorentz transformation/Archive 5.

Best, 'M'&and;Ŝc2ħεИτlk 18:59, 13 December 2015 (UTC)


 * Thank you for thinking about me, but really I don't know enough to help you with your page. Maybe better to ask some of the editors of hyperboloid model or Minkowski space or put the questions you have at or  hope this helps, I am not sure even if the model is quantitative enough for hyperbolic geometry, good luck  WillemienH (talk) 20:06, 14 December 2015 (UTC)


 * No worries ans thanks for replying ^_^ Also, it's not my page! 'M'&and;Ŝc2ħεИτlk 22:13, 15 December 2015 (UTC)

Detailed contents of Euclids elements Book 1
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Speedy deletion nomination of Detailed contents of Euclids elements Book 1


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Doxastic logic
I dispute that the proof you corrected was invalid. :)

The only step I notice you rejecting went from ℬ((ℬS → S) → S) and ℬ(ℬS → S) to ℬS, but a type 1 reasoner's beliefs are logically closed under MP, meaning by definition that he/she will infer q from p→q and p, so it's okay to infer ℬq from ℬ(p→q) and ℬp. The axiom ℬ(p→q) → (ℬp → ℬq) ensures this.

(My earlier proof was invalid in the sense that it used (ℬp ∧ ℬq) → ℬ(p∧q) — this is called agglomeration — which is often included, but isn't in this treatment, and might be worth a mention as an optional additional rule. But I caught that issue myself upon further review.)

JLM~enwiki (talk) 06:52, 3 January 2016 (UTC)


 * Maybe it is just the way I do proofs, I like to follow the very stict rules of Formal proof "A formal proof or derivation is a finite sequence of sentences (called well-formed formulas in the case of a formal language) each of which is an axiom, an assumption, or follows from the preceding sentences in the sequence by a rule of inference. ...".


 * If you look very strict there is no rule of inference that allows you to go from ℬ((ℬS → S) → S) and ℬ(ℬS → S) to ℬS. There is only given that for a reasoner : "If he or she ever believes p and believes p → q (p implies q) then he or she will (sooner or later) believe q " as an extra valid formula scheme (ℬp & ℬ(p → q)) →ℬq. It is important to realise this is not a rule of inference or an tautology but only an extra valid formula. (These are very fine lines, if in doubt use it only as valid formula) so if you want to use it in your proof it needs to be mentioned. not mentioning it gives an invalid proof.


 * Having said all this, you can argue that I am still a bit lax with the inference rules.


 * S ≡ ¬&#x212C;S [definition of S]
 * (¬S → S) → S [elementary tautology]
 * (&#x212C;S → S) → S [because ¬S ≡ &#x212C;S]

Should be replaced with:


 * 1 S ≡ ¬&#x212C;S [definition of S]
 * 2 (¬S → S) → S [elementary tautology]
 * 3 (¬S ≡ ¬&#x212C;S ) → (((¬S → S) → S)→ (&#x212C;S → S) → S [(not so) elementary tautology]
 * 4 (((¬S → S) → S)→ (&#x212C;S → S) → S [modus ponens 1/3]
 * 5 (&#x212C;S → S) → S [modus ponens 2 /4 ]


 * and you would have a fair point. Good luck WillemienH (talk) 09:23, 3 January 2016 (UTC)


 * It strikes me as excessive mincing to disallow treating a valid implication as an inference rule, and require using a tautology to get the expression into the form used by the formal rule unless logics which lack Conditional Proof or otherwise make inference by implication not universally valid are in the realm of discourse. Regarding laxity in your "correction", I'd consider your use of statements p and q to satisfy p∧q without including the statement p∧q as a line in the proof to be a bigger sin than admitting substitution of equivalent subexpressions (i.e., it's nitpicking which doesn't actually invalidate your proof, like I feel yours doesn't mean my proof is incorrect).  Anyway, I added a note to the rule on type 1 reasoners about how (a∧b)→c is equivalent to a→(b→c) in the main article, so I hope that'll clarify things without having to double the length of the proof and thereby obscure the reasoning behind it. JLM~enwiki (talk) 18:43, 3 January 2016 (UTC)


 * Just be very carefull. In no time you will be able to proof S → &#x212C;S and then you have lost the plot :) ( S → &#x212C;S is not a valid formula and therefore should not be provable). and indeed under very strict interference rules  Conditional Proof is not a valid interference rule. (mostly happens in logic that have the nessRule of necessitation, and doxastic logic is a logic where this rule can be used. WillemienH (talk) 19:24, 3 January 2016 (UTC)


 * Ho, but I think I can!
 * 1. ¬S [Result of WillemienH 30-Dec-2015 ]
 * 2. ¬S ∨ ℬS [disjunction introduction on 1]
 * 3. S → ℬS [material implication on 2]
 * JLM~enwiki (talk) 20:55, 3 January 2016 (UTC)
 * JLM~enwiki (talk) 20:55, 3 January 2016 (UTC)

Sorry, I mean proof that S → &#x212C;S is a theorem (so without assuming anything of the previous proof) something like:
 * 1. S [assumption]
 * 2. ℬS [1 Rule of necessitation]
 * 3. S → ℬS [conditional proof discharging assumption 1]

no this deduction is not valid, but


 * 1. S → S [assumption]
 * 2. ℬ(S → S) [1 Rule of necessitation]
 * 3. (S →S) → ( ℬ(S→S)) [conditional proof discharging assumption 1]

is valid, can you see why? WillemienH (talk)


 * Yes, my "proof" of S → ℬS was a joke. I used the meaning of S from the main article where it's defined to be ¬ℬS, but you obviously (though implicitly) meant it to be an arbitrary proposition with implicit universal quantification.  I should have put in a second smiley!
 * Since you love to split hairs, your 2nd example immediately above is also incorrect: Though the statements proper match a valid usage of necessitation, their structure is such that necessitation is improper to apply where you do, because you've chosen to take S→S as an assumption and use conditional proof.  If you had eschewed conditional proof and taken S→S as a tautology instead, you would have had a valid proof as long as you accept q ⇒ x→q as an inference rule:
 * 1. S→S [tautology]
 * 2. ℬ(S→S) [necessitation]
 * 3. (S→S) → ℬ(S→S) [q ⇒ x→q with q=ℬ(S→S) and x=(S→S)]
 * JLM~enwiki (talk) 22:58, 3 January 2016 (UTC)

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Update on images of quadrics
Hey. Just to let you know I've updated the images on the article quadric as you requested. I couldn't match the colour scheme exactly, but I hope it's still OK. Thanks again for the request. —SamTalk 14:41, 12 April 2016 (UTC)

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