A counterexample for equivalency of $P \implies Q$ and $\neg Q \implies \neg P$? Suppose P is when it rains and Q is when a garden's soil is dry. Now, I cover the garden with a cover. It rains. So I say a true statement: If it rains the garden's soil is not (won't be) wet. And its contrapositive is: If the garden's soil is wet so it didn't rain. But it rained!
Is this a counterexample for $P \implies Q$ and $\neg Q \implies \neg P$? I don't know about logical symbols neither I don't know Logic; simple clear explanations would be much appreciated.           
 A: You end your discussion with the situation where the garden's soil is wet and it rained.  Therefore, $P$ is true and $\neg Q$ is also true.  This contradicts your original assumption that $P\Rightarrow Q$ is true.
The only way for $P\Rightarrow Q$ to be true is that whenever it rains, it is impossible for the soil to be wet (even, perhaps, for other reasons).  In other words, for $P\Rightarrow Q$ to be true, then when it rains, the rain prevents all possible reasons for the soil to be wet (it's not just that the soil is not wet because of the rain).
Try not to confuse causality with implication.  $P\Rightarrow Q$ means that whenever $P$ is true, then $Q$ must also be true; this does not mean that $P$ causes $Q$.  Think about this case (you have a class with two exams and no other graded work): $A$ is "I failed the class" and $B$ is "I failed an exam."  $A\Rightarrow B$ is true because if you failed the class, then you had to have failed an exam (note that if you fail an exam, you might not have failed the entire class, so $B\Rightarrow A$ is false).  In this case, the observation of $A$ allows you to conclude that $B$ happened, but $A$ does not cause $B$.  In fact, $B$ partially causes $A$, but $B$ does not imply $A$.
Therefore, neither $P\Rightarrow Q$ nor $\neg Q\Rightarrow \neg P$ are true in this case.  
In order to fix your original statements, let $P$ be the event that it rains and $\neg Q$ is the event that the garden is wet because of rain.  (Then $Q$ is the garden is not wet or the garden is wet for some reason other than rain.)  Then $P\Rightarrow Q$ and $\neg Q\Rightarrow \neg P$ are true because in your final consideration, $\neg Q$ is false, the garden is wet, but not because of rain.
A: When we say $P\to Q$ it means that $Q$ happens to be true whenever $P$ is.
This means either $Q$ is true or $P$ is false. 
This means $P$ happens to be false whenever $Q$ happens to be false. 
$$P\to Q
\\ \Updownarrow
\\ \neg P\vee Q
\\ \Updownarrow
\\ \neg Q\to\neg P$$

When you say "If it rains, then the soil won't be wet", you are giving a guarantee that (somehow) the soil will not be wet whenever it rains.   So assuming your guarantee is holds: it does not rain whenever the soil is wet would be a truth.
If ever the soil is wet and it rained, then your guarantee didn't hold.   Neither the statement $P\to Q$ nor the contrapositive $\neg Q\to \neg P$ would be true.   In that situation, they both would be false at the same instance; hence this is not a counter example against their equivalence.


Logical equivalency of $P⇒Q$ and $¬Q⇒¬P$ is something that I can't understand it 'directly' and I have to use many examples. Could you please help me with a way to understand it with reasoning not by examples? Thank you. – Liebe 

Do you grok truth tables?
$$\begin{array}{|c:c|c|c|c:c|c|}\hline
~P~ & ~Q~ & ~ P\to Q ~& & \neg Q~ & \neg P~ & \neg Q \to \neg P~
\\ \hline
\color{blue}T & \color{blue}T & \color{blue}T & & \color{red}F & \color{red}F & \color{blue}T
\\ \hdashline
\color{blue}T & \color{red}F & \color{red}F & & \color{blue}T & \color{red}F & \color{red}F
\\ \hdashline
\color{red}F & \color{blue}T & \color{blue}T & & \color{red}F & \color{blue}T & \color{blue}T 
\\ \hdashline
\color{red}F & \color{red}F & \color{blue}T & & \color{blue}T & \color{blue}T & \color{blue}T
\\ \hline
\end{array}$$
