User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

Tarski's Axioms of Geometry
I'm going to have to figure out how to precisely and accurately express what are the disciplines preceding Tarski's geometry. I'll post my progress here for anyone who wants to follow.

GFP to Steven Givant:

If you have time, I'd like to ask you a question about your essay.

...

Is this to say that Aristotelian logic precedes Tarski's Geometry? That is, does Tarski's system depend on the law of excluded middle, the principle of non-contradiction, modus ponendo ponens, etc. etc. ? -GFP

Stevent Givant to GFP:

Tarski assumes the language and axioms of first -order logic (which is substantially stronger than propositional logic---what you might be referring to when you speak of Aristotelian logic). - Givant

Comments are welcome. Does PW have a page on the order of a logical statement? I couldn't find one. --GFauxPas 15:00, 25 January 2012 (EST)


 * First-order logic is stronger than propositional logic as it allows quantification ($\forall, \exists$). The proofs of Tarski-Vaught Test and Łoś's Theorem implicitly use an order on formulas. This however has not been properly defined on PW. So go ahead :). --Lord_Farin 15:19, 25 January 2012 (EST)

I'm working on the axiom of continuity and could use some help, please.

The axiom schema is:


 * $\exists a: \forall x,y: \left({ \alpha \land \beta \implies \mathsf{B}axy }\right)$


 * $\implies \exists b: \forall x,y: \left({ \alpha \land \beta \implies \mathsf{B}xby }\right)$

where $\alpha,\beta$ are first order formulas where $\alpha$ doesn't contain free occurances of $a,b,y$ and the second doesn't contain free occurances of $a,b,x$.

So we're making sure the "line" is "full". But why is it important that $\alpha,\beta$ not contain free occurances of the quantified variable? Wouldn't they automatically become bound by $\exists \forall$, and it would "fix itself"? And why aren't we worried about bound occurances? :( This is different from the zeroth and first order logic I'm used to.

This seems similar to Axiom:Density of Betweenness, but more so. --GFauxPas 12:14, 26 January 2012 (EST)


 * You say: Wouldn't they automatically become bound by $\exists \forall$, and it would "fix itself"?. They would indeed become bound. However, the meaning of the expression $\alpha\land\beta$ would (or at least, could) become different. The idea is that $\alpha$ says something about $x$, similar to $x \in X$, and $\beta$ says something about $y$, like $y\in Y$. I can't produce an example at this time, but that's what the catch is about. --Lord_Farin 13:07, 26 January 2012 (EST)

Okay I need help I have no idea what to call these axioms:

A23: If line segment xy is attached to line segment yz, and line segment $x'y' \equiv y'z'$, and $yz \equiv y'z'$, then line segments $xyz$ and $x'y'z'$ are the same length.

A24: If you have two line segments the same length, and you cut off the same length of each, the remaining parts will be the same length.

I'm using an intuitive description here in the sandbox in the hope it will be more suggestive of a title. --GFauxPas 12:48, 26 January 2012 (EST)


 * What I came up with is something with 'additivity', similar to Definition:Additive Function (Measure Theory). For example, 'First Additivity Axiom for Equidistance' and 'Second ...'. The analogy fits ('the whole is exactly the sum of the parts'). --Lord_Farin 13:07, 26 January 2012 (EST)
 * Looking at the naming so far, 'Outer...' and 'Inner...' may also be a good choice for constistence. --Lord_Farin 13:08, 26 January 2012 (EST)


 * $u \ v \ \mathsf{u} \ \mathsf{v} \ \nu \ \upsilon$

Anyone else have a hard time distinguishing between $u$ and $v$? I would like it to look more like this, does it confuse anyone else? It seems PW doesn't have the upgreek package. --GFauxPas 07:49, 27 January 2012 (EST)


 * Nope. Multiple years of extensive TeX writing and reading have trained my eye. I agree that referenced $v$ looks more distinguished, but imagine it is hard to implement. --Lord_Farin 08:08, 27 January 2012 (EST)