# User talk:Quondum

## The least element of the subset ω+Z:

I've thought again about your question, and here is the correct answer: What is only provable by First order Axiom of induction is about the least element satisfying a given First order proposition, rather than about the least element belonging to a given subset. So, really your subset ω+Z has no minimum, but First order Peano language can formulate no proposition φ(x) satisfied only by every x belonging to your subset. HOTmag (talk) 12:06, 9 August 2016 (UTC)

I'm a rookie with the Peano axioms, but I was just trying to point out what seemed to me to be an error in the reasoning. Identifying such an error does not require the rest of the system to be understood. The apparent error involved concluding that one could prove a statement that was expressible as a first-order statement (that there existed an element that satisfied a given proposition) using a wellordering. However, my reasoning was not followed, so it seemed best to exit the discussion.
I don't quite follow what you say above (the logic is not clear to me), but I suspect that your conclusion (that no proposition that identifies nonstandard elements exist) is true. —Quondum 15:13, 9 August 2016 (UTC)
I was referring to what you'd said here: "Okay, so how is the well-ordering principle a consequence of the axiom of induction? My example of a nonstandard model seems to an example that violates the principle, but does not violate the axiom of induction. To recap: The well-ordering principle says that every subset of the model has a least element. My model has a subset, namely ω+Z, that has no least element under the ordering <. Thus, < is not a well-ordering in this model ".
So I explained that your model, which has a subset (namely ω+Z) having no least element under the ordering <, does not disprove the fact that the well-ordering principle is a consequence of the First order Axiom of induction, because the "consequence" of First order Axiom of induction is about the least element satisfying a given First order proposition, rather than about the least element belonging to a given subset. So, really your subset ω+Z has no minimum, but First order Peano language can formulate no proposition φ(x) satisfied only by every x belonging to your subset. HOTmag (talk) 16:52, 9 August 2016 (UTC)
So as I synthesize this, you are saying that since no first-order proposition can exist that is true exactly for nonstandard elements, we cannot have a proposition that switches truth-value "between" the nonstandard and the standard elements. And thus there must be a least value at which it does switch, whether this is at a standard or nonstandard element. (I still have to think through the premise here, but I'm not arguing against it.) —Quondum 18:26, 9 August 2016 (UTC)
Your first sentence ("since no...standard elements.") is correct. However, your second sentence ("And thus there must be a least value...nonstandard element.") is not clear to me, because the "least value" I'm talking about is related to a first order proposition (as I've explained many times) - whereas you mentioned no first order proposition when you mentioned the "least value". HOTmag (talk) 19:08, 9 August 2016 (UTC)
I meant "And thus there must be a least value at which it [the truth value of any given first order proposition] does switch ...". But I have also been assuming by "least" that a total ordering < is well-defined, and matches our perception of it (in particular that if we somehow define +, a<b ⇔ ∃x(a+S(x)=b), even though for my logic < need not itself be expressible as a first-order preposition. I may be glossing over similar points due to me making assumptions without realizing it. —Quondum 19:57, 9 August 2016 (UTC)
The order < must satisfy: a<b ⇔ ∃x(a+S(x)=b), and it's also well defined by that first order definition. HOTmag (talk) 09:55, 11 August 2016 (UTC)
I'm not sure which definition you are referring to when you say "that first order definition". Nevertheless, my original point was that this < is not a wellorder, and hence the wellorder principle does not seem to apply. Remember, I was arguing about the logic, not the conclusion. Anyhow, I'm not sure where we're headed with this: my knowledge of the area is very poor as I've already pointed out, and until I study a lot of the basics in this area, I am liable to tripping over things. —Quondum 16:13, 11 August 2016 (UTC)
I was referring to the definition I'd just mentioned, i.e. a<b ⇔ ∃x(a+S(x)=b), and this order < - defined by that first order definition - is a wellorder in the following sense: Every first order formula φ(x) with x as a free variable, has (under < ) a least element x satisfying φ(x). Further, this fact is a consequence of the axiom of induction. That's what I've been claiming from the very beginning, and I wonder what isn't clear yet - even to you (and to everybody who has elementary knowledge in logic, even without knowledge in Peano axioms). HOTmag (talk) 17:19, 11 August 2016 (UTC)
We are clearly running into communication clarity issues. For example, something introduced with the words "must satisfy" does not qualify as a definition, yet you've just referred to it as a definition. We clearly have a number of these little disconnects to settle before we both will be using the same terminology. Let's discontinue this. —Quondum 18:04, 11 August 2016 (UTC)
Just for the sake of clarity: By "must satisfy" I meant, that since the order < is defined (in Peano system) by the statement: a<b ⇔ ∃x(a+S(x)=b), then (as a result) it must satisfy that formula (as far as Peano system is concerned). In other words: not only must the order < satisfy that formula, it is also defined by that formula. Sorry for not making it clear from the very beginning. HOTmag (talk) 19:05, 11 August 2016 (UTC)