Logic of the implication in $ε$-$δ$ proofs Im confused by why epsilon delta proofs logically work.
An example is 
Proof: 
Given $ε>0$, choose $δ = {ε\over3}$. For all $x$, if $0<|x−2|<δ$ then $|(3x−1)−5| < ε$.
That last part  if $0<|x−2|<δ$ then $|(3x−1)−5| < ε$ LOOKS A LOT LIKE $P\to Q$ because of the "if then" but yet the proof in the book solves it like as if its $Q\to P$?:
$$\begin{align}|(3x−1)−5| &= |3x−6|\\ &= |3(x−2)|\\ &= 3|x−2|\\ &<3δ\\ &= 3\left({ε\over3}\right) \\ &= ε\end{align}$$
So my question is how come it looks like a $P\to Q$ proof but yet we start with $Q$ to show $P$?

In the proof we do not start with $Q$, namely, we don't assume that  $|(3x−1)−5| < \varepsilon$. If you read the string of manipulations carefully, you see that 
$$
|(3x−1)−5| = |3x−6| = |3(x−2)| = \color{blue}{3|x−2|<3\delta= 3(\varepsilon/3) } = \varepsilon.
$$
we use the assumption that $0<|x-2|<\delta=\varepsilon/3$ in the steps coloured blue to show that $|(3x−1)−5|<\varepsilon$.
