Problem:
Show that these three statements are equivalent, where $a, b \in R:$ (i) $a < b$, (ii) the average of $a, b,$ is greater than $a,$ and (iii) the average of $a$ and $b$ is less than $b$.
(Note: to show that three statements are equivalent it is enough to show that (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i); why? Because of hypothetical syllogism!)
Question:
Do I need to set up implication proofs (i.e. (i)$\rightarrow$(ii)$\equiv\neg$(i)$\lor$(ii)$\equiv$ etc.) between each relation or is the better way to prove the statements are equivalent to show how to transform one statement to the next? (see example below)
Here is what I have: (is this the best way to complete the problem?)
Proof. Suppose it is true that $a<b$.
Then add $a$ to both sides of $a<b$ and solve for $a$.
$$a+a<a+b \equiv 2a<a+b \equiv a<\frac{a+b}{2}$$
Therefore $(a<b)\Rightarrow(a<((a+b)/2))$.
Then add $b$ to both sides of $(a<((a+b)/2))$ and solve for $b$.
$$a+b<\frac{a+b}{2}+b \equiv a+b-\frac{a+b}{2}<b \equiv \frac{a+b}{2}<b$$
Therefore $a<((a+b)/2))\Rightarrow((a+b)/2))<b$.
Then multiply both sides of $((a+b)/2))<b$ by $2$ and solve for $a$.
$$2\frac{a+b}{2}<2b \equiv a+b<b+b \equiv a<b$$
Therefore $((a+b)/2))<b\Rightarrow a<b$.
Consequently $(i)\Rightarrow(ii)\Rightarrow(iii)\Rightarrow(i)$.
Therefore $(i)$,$(ii)$, and $(iii)$ are equivalent.
Yes, Hypo Syll works this way :
You can use it, plus the fact that $P \equiv Q$ is $(P \rightarrow Q) \land (Q \rightarrow P)$.
When you haved "closed the cricle", having proved that (iii) $\Rightarrow$ (i), you will have :
(ii) $\Rightarrow$ (iii), so with the above, apply HS and you will have (ii) $\Rightarrow$ (i), that added to (i) $\Rightarrow$ (ii) gives you : (i) $\equiv$ (ii).
The same applies with (i) $\Rightarrow$ (ii) and (ii) $\Rightarrow$ (iii), that gives you (again using HS) : (i) $\Rightarrow$ (iii), to be added to (iii) $\Rightarrow$ (i) to have : (i) $\equiv$ (iii).
Then, by properties of $\equiv$, you have also : (ii) $\equiv$ (iii).