Do we have to check the second case?

[evinda]

Hi! (Smile)

Proposition:
The natural number $n'$ is the immediate successor of $n$, i.e. there is no natural number $m$ such that $n<m \wedge m<n'$.

Proof:
We assume that there is a $m$ such that $n<m \wedge m<n'$. Then $n \subset m$ and $m \subset n \cup \{n\}$. 1st case: $n \in m$. Then $n \cup \{n\} \subset m$ and so $n'=m$, contradiction.

2nd case: $n \notin m$. Then since $m \subset n \cup \{n\}$ we have that $m \subset n$. Since $n \subset m$ we have $n=m$, contradiction.
Do we have to check the second case knowing that $n<m \leftrightarrow n \in m$ ? (Thinking)

 

[Evgeny.Makarov]

Do we have to check the second case knowing that $n<m \leftrightarrow n \in m$ ?

No, if $n<m \leftrightarrow n \in m$ is proved before the current proposition, then the second case is unnecessary.

 

[evinda]

No, if $n<m \leftrightarrow n \in m$ is proved before the current proposition, then the second case is unnecessary.

It hasn't been proven, it is referred in a proposition. (Worried)
How could we prove it? (Thinking)

 

[Evgeny.Makarov]

It hasn't been proven, it is referred in a proposition.

What does this mean: "it is referred in a proposition"?

How could we prove it?

You should start with a definition of $<$. Isn't this statement proved in your lecture notes?

 

[evinda]

According to my notes:

Definition: For any natural numbers $m,n$ we define:

$m<n \leftrightarrow m \in n (\leftrightarrow \langle m,n \rangle \in \epsilon_{\omega})$

$(m \leq n \leftrightarrow m \in n \lor m=n)$​

where $\epsilon_{\omega}=\{ \langle m,n \rangle\in \omega^2: m \in n \}$.$\epsilon_{\omega}$ defines an order on $\omega$ and is symbolized with $<$.
The non-strict order of $\omega$ that corresponds to the order $\epsilon_{\omega}$ is symbolized with $\leq$, so:

$$n \leq m \leftrightarrow n \in m \lor n=m$$
$$(n<m \leftrightarrow n \in m)$$

Proposition:

• For any natural numbers $m,n$: $m \leq n \leftrightarrow m \subset n$
• For any natural numbers $m,n$: $m<n \leftrightarrow m \subsetneq n (\leftrightarrow m \in n)$

 

[Evgeny.Makarov]

No, if $n<m \leftrightarrow n \in m$ is proved before the current proposition, then the second case is unnecessary.

How could we prove it?

According to my notes:

Definition: For any natural numbers $m,n$ we define:

$m<n \leftrightarrow m \in n$​

Then you answered your own question.

 

[evinda]

So we can use the fact that $m<n \leftrightarrow m \in n$ without proving it since we defined it like that, right? (Smile)

 

[Evgeny.Makarov]

Yes.

 

[evinda]

Nice... Thanks a lot! (Yes)