Hello everybody:) I have a question, hope you could tell me the answer. What is the gist of Soundness Theorem?

@TungNguyen never heard of such rule

@Michelle you mean the one that says $(T \vdash \varphi) \implies (T \vDash \varphi)$?

The soundness theorem generally refers to the converse of the completeness theorem

yes @LeakyNun

by "gist" you mean "point"?

It means if you start with true assumptions, and apply the rules of inference, you will end with a true statement

like the answer would be "soundness theorem essentially says that ..."?

or what do you mean by "gist"?

for instance, like if I wanted to explain the main idea of the proof

not the formal proof

the proof of the theorem or the theorem itself?

the proof

sorry I wasn't clear with my question.

by induction on the length of proof, as my professor likes to say

You basically show what leaky has typed in latex above holds for each rule of inference

it all depends on the rules you actually have

I just have 2 rules, MP and generalization

it actually also depends on the rule for something to be a structure

MP is ingrained in the definition of structure

i.e. every structure has to satisfy MP

(you might call a structure a "valuation" depending on what you follow)

It will depend on virtually every aspect of the logical system you have developed. Both the method of proving things (as different systems will have different notions of proof) and the semantics of your system (how you have defined what it means for a sentence in your system to be "true")

when it's a theorem or when it's a valid formula under any interpretation

both

yes

In other words, you prove it for any set of assumptions, including the empty set (tautologies/valid statements)

@Michelle The best person to ask these types of questions to is @user21820 :) As he literally holds logic classes in here and will lock you in here until he is convinced you understand it better than he does ;)

He's in here every day, should be logging in within the next 3 hrs or so

He is familiar with virtually every logical system there is and will likely be able to give you better feedback suited to the specific system you are using

okay

@DavidReed So accurate!

@user21820 :o

@TungNguyen They are not equal. They just so happen to be equivalent. Let's put it another way. "1=2" is equivalent to "1=3", not because their logical forms are inherently equivalent, but simply because we can prove both false. Similarly, there are some sentences that are equivalent not because they are instances of some general equivalent pairs, but just because we can prove them equivalent.

@LeakyNun I was indeed here about an hour ago, but haven't come to this room yet.

@TungNguyen: Did you also read and understand my response starting here?

@Michelle In general, a system that involves rules typically means that each rule can be thought of as an operation on valid statements; when you apply the rule to existing statements, it gives more statements. The theorems of the system are then generated by applying the rules to all the axioms, and keep applying them to all newly generated statements, and so on until the 'end'. Mathematically we simply say that the theorems are the closure of the axioms under the rules.

In general soundness of a system means that every theorem it proves is true in every model of that system. For rule-based systems I just described, it suffices (but may not be necessary) to prove that each rule is sound, namely when you apply it to true statements it gives only true statements. We call such rules truth-preserving.

So you can summarize the proof of the soundness theorem by saying that you prove that every rule is truth-preserving, and hence the closure of the axioms under the rules only includes true statements, since every statement in the closure is generated by finitely many applications of truth-preserving rules.

@LeakyNun @DavidReed: Did I miss out any important point?

Negative, I think she was interested in the particulars for a system that consisted only of cut/MP and quantifier introduction

Negative as in I don't think you missed anything, not negative as in wrong :)

@DavidReed If i'm not wrong she's learning from a book that uses a purely Hilbert-style system, so it does not even have cut/quantifier-rules.

→ 8 messages moved to trash

She said "MP and generalization"

to be more precise :)

@DavidReed Oh okay I missed that. In many textbooks they would just use axiom schemas to capture "generalization", and so MP would be the only rule. Convenient for meta-theorems, but terrible for actual human use.

> it betrays muddleheadedness

challenge: come up with a more pretentious way of speaking

> But this is a $\omega_1$ feat. Do you think that what is open must be closed?

> (for i < $\omega$, output = Nonsense. If uud=#787, then) > dev/null

> You idea is entirely $\omega_1$, which means we need to accept the axiom of choice for it to be trivial

@LeakyNun Challenge accepted:

> it is the epitomy of having a pudding for a head

@LeakyNun: Did you find any problem with my recent attempt?

← 3 messages moved from Mathworks (Not the main chat!)

Where R = ( func(nat,obj) f ↦ ( nat n ↦ f(n+1) ) ), and the above axioms hold for every f∈func(nat,C*).

Oh the "\{1}" is not necessary, since I defined 1/0 = ∞. I should also say ∞/∞ is undefined.

hi , i have a question:

L is a language with a structure M, s.t for each $a\in |M|$ there is a constant $c_a $ in $L$ s.t $c_a \ ^ M = a$

i need to prove that $Th(M)$ satisfy henkin condition

someone can help?

@LeakyNun maybe you?

not me

@Liad that's almost an immediate consequence so... What's your statement of the Henkin condition?

@AlessandroCodenotti i thought so too. the definition is that if $A$ is a closed formula in the form $\exists x B$ then there is a constant $c$ s.t $T \vdash \exists x B \to B_x[c]$

Ok, so pick a sentence $A$, if $\text{Th}(M)\vdash \neg \exists x B$ the Henkin condition is satisfied, what if $\text{Th}(M)\vdash\exists x B$? (I'm using the fact that $\text{Th}(M)$ is complete here)

@AlessandroCodenotti sorry didnt saw you answered. if $T \vdash A$ then for all $\sigma :Vars \to \{ \Bbb T , \Bbb F\}$ ,$\sigma(A) = \Bbb T$ so there is $a\in |M|$ s.t $\sigma(B) = \Bbb T$

i want to say this implies that $T\vdash A \to B_x[c_a]$

(@AlessandroCodenotti you also need to assume $A$ is closed)

i needed to write "if $M \vDash A$ " not "$T \vdash A$"

It's the same thing, the theory of $M$ proves exactly the sentences that $M$ models

@Liad Ok, so this $a$ is the interpretation of some constant by assumption

@AlessandroCodenotti but why $T \vdash A \to B_x[c_a]$ ?

we only have $\sigma(x/a)(B) = \Bbb T$

That's the same as asking $M\vDash A\rightarrow B_x[c_a]$, right? And $T\vdash A$ also means that $M\vDash A$

@AlessandroCodenotti dont you need that every structure will satisfy $M' \vDash A\to B_x[c_a] $ in order to conclude that $T \vdash A \to B_x[c_a]$ ?

every structure that satisfy $Th(M)$

You have $T=\text{Th}(M)$ here, right?

yes

That theory is complete so any two models are elementarily equivalent, if you have $A,B\vDash T$, then for all sentences $\varphi$, $A\vDash\varphi\iff B\vDash\varphi$

You are right. i forgot about the definition of complete theory.. thanks! @AlessandroCodenotti

@LastIronStar @Secret: I would like to inform you that user 170039 has recently removed my write-access to the Philosophy chat-room, and has also supported George Chen here, despite knowing that he incites violence and knowing that he posted junk all over Math SE.

A lot of nonsensical comments on that Meta thread have been deleted by the moderators, but many of those from George were recorded in my post, and one more later one here.

*Philosophy of Mathematics chat-room.

See also user 170039's false claim here.

I feel it is necessary to warn you about all this, since you may not be aware of what has happened. Moreover, note that the claim by George that Godel's sentence is self-referential is false and clearly reveals that he in fact does not understand the incompleteness theorem at all. Note the irony that he tells people to give Godel's work a finishing blow, despite not having understood it at all.

@user21820 I'm curious as to how you would solve $f'(x) = f(x)$ (rigorously)

if you want the formal statement: find all possible functions $f:\Bbb R \to \Bbb R$ that is differentiable everywhere such that $f'(x) = f(x)$ for all $x \in \Bbb R$