Proof of Second Sentence:$m<n \leftrightarrow m'<n'$

[evinda]

Hello! (Wave)

I am looking at the following sentence:

For any natural numbers $m,n$ it holds:

• $m \leq n \leftrightarrow m' \leq n'$
• $m<n \leftrightarrow m'<n'$
• $m<n' \leftrightarrow m \leq n$
• $m \leq n' \leftrightarrow m \leq n \lor m=n'$

I tried to prove the second sentence like that:

$$m<n \rightarrow m \in n \rightarrow m \subset n \wedge \{m\} \subset n \rightarrow m \cup \{ m \} \subset n \rightarrow m \cup \{m \}=n \lor m \cup \{ m \} \in n$$

From the relation $m \cup \{m \} \in n$ we get that $m \cup \{ m \} \in n \cup \{ n \} \rightarrow m' \in n' \rightarrow m'<n'$.

$$m'<n' \rightarrow m' \in n' \rightarrow m \cup \{ m \} \in n \cup \{ n \} $$

$$m \in m \cup \{ m \} \rightarrow m \in n \cup \{ n \} \rightarrow m \subset n \cup \{ n \} \rightarrow m \subset n \lor m \subset \{ n \}$$

From the relation $m \subset n$ we conclude that $m \in n \lor m=n$.

How could we reject the case $m=n$ ? (Thinking)

 

[Evgeny.Makarov]

$$m'<n' \rightarrow m' \in n' \rightarrow m \cup \{ m \} \in n \cup \{ n \} $$

$$m \in m \cup \{ m \} \rightarrow m \in n \cup \{ n \} \rightarrow m \subset n \cup \{ n \} \rightarrow m \subset n \lor m \subset \{ n \}$$

From the relation $m \subset n$ we conclude that $m \in n \lor m=n$.

How could we reject the case $m=n$ ?

You could argue that if $m=n$, then $m'=n'$ and therefore $m'<n'$, i.e., $m'\in n'$, is impossible.

 

[evinda]

You could argue that if $m=n$, then $m'=n'$ and therefore $m'<n'$, i.e., $m'\in n'$, is impossible.

I see! (Nod)

Could we show the fourth proposition ($m \leq n' \leftrightarrow m \leq n \lor m=n'$) like that? (Thinking)

$$m \leq n' \leftrightarrow m=n' \lor m \in n' \leftrightarrow m=n' \lor m \in n \cup \{ n \} \leftrightarrow m=n' \lor m \subset n \cup \{ n \} \\ \leftrightarrow m=n' \lor m \subset n \lor m \subset \{ n \} \leftrightarrow m=n' \lor m \leq n \lor m \subset \{ n \}$$

It holds that:

$$m \subset \{ n \} \rightarrow m=\{ n \} \lor m \in \{ n \} \rightarrow m=\{ n \} \lor m=n$$

But we know the following:

$$m \leq n \leftrightarrow m \in n \lor m=n$$

and:

$$\{ n \} \subset n$$

So, $m \subset \{ m \} \rightarrow m \leq n$.

Therefore,

$$m \leq n' \leftrightarrow m=n' \lor m \leq n.$$

 

[evinda]

Hello! (Wave)

I am looking at the following sentence:

For any natural numbers $m,n$ it holds:

• $m \leq n \leftrightarrow m' \leq n'$

I tried to prove the second sentence like that:

$$m<n \rightarrow m \in n \rightarrow m \subset n \wedge \{m\} \subset n \rightarrow m \cup \{ m \} \subset n \rightarrow m \cup \{m \}=n \lor m \cup \{ m \} \in n$$

From the relation $m \cup \{m \} \in n$ we get that $m \cup \{ m \} \in n \cup \{ n \} \rightarrow m' \in n' \rightarrow m'<n'$.

We say that $m \leq n \leftrightarrow m \in n \lor m=n$ but then we only use the case when $m \in n$.. Do we also have to conclude something from $m=n$?

 

[blue_raver22]

As a scientist, it is important to consider all possibilities and not reject any potential case without sufficient evidence. In this case, we cannot reject the possibility that $m=n$ without further information or context. It is possible that in certain scenarios, $m=n$ may still hold true and does not necessarily contradict the given proof. It is important to keep an open mind and continue exploring the possibilities until all evidence has been thoroughly evaluated.