Does it suffice to show these relations?

[evinda]

Hi! (Wave)

If I want to prove that $A \cap B=A \text{ iff } A \subset B \text{ iff } A \cup B=B$.
Do I have to prove the following:
$A \cap B=A \rightarrow A \subset B$, $A \subset B \rightarrow A \cap B=A, A \subset B \rightarrow A \cup B=B, A \cup B=B \rightarrow A \subset B $ and $A \cup B=B \rightarrow A \cap B=A$ ?

 

[Evgeny.Makarov]

To show $P\iff Q\iff R$ it is sufficient to prove, for example, $P\implies Q$, $Q\implies R$ and $R\implies P$. At least three implications are necessary, but they can be chosen in different ways.

 

[evinda]

To show $P\iff Q\iff R$ it is sufficient to prove, for example, $P\implies Q$, $Q\implies R$ and $R\implies P$. At least three implications are necessary, but they can be chosen in different ways.

So, don't we have to show, for example, $Q \Rightarrow P$ ? (Thinking)

 

[magneto1]

So, don't we have to show, for example, $Q \Rightarrow P$ ? (Thinking)

If you can show $Q \Rightarrow R$ and $R \Rightarrow P$, that immediately implies $Q \Rightarrow P$; it is unnecessary to show it explicitly.

 

[evinda]

If you can show $Q \Rightarrow R$ and $R \Rightarrow P$, that immediately implies $Q \Rightarrow P$; it is unnecessary to show it explicitly.

A ok.. But, if I would prove also $Q \Rightarrow P$, would it be wrong?

 

[magneto1]

A ok.. But, if I would prove also $Q \Rightarrow P$, would it be wrong?

It is not wrong. You can show the implications in any order: E.g $Q \Rightarrow P \Rightarrow R \Rightarrow Q$, or $R \Rightarrow P \Rightarrow Q \Rightarrow R$.

In fact, you usually want to choose an ordering that makes the proof the simplest if possible.