MHB 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 $m=n$?