If $(P \implies Q),$ then $\lnot(P\land\lnot Q)$ In this correct:

If $(P \implies Q)$ is true, then $\lnot(P\land\lnot Q)$ is true. 

I came up with this in my search to understand implication and the two troublesome, for me, lines in the truth table.  I am only able to write this sentence, I do not have the knowledge to prove it - yet. I can show that the truth table columns for two statements are the same.

This is either my first glimmer of light or my first false step.

I wrote "¬(P∧¬Q) is true" to describe both the line in the truth table where P is true and Q is false and further explore the implication connective.  I thought that "if (P⟹Q) is true, then ¬(P∧¬Q) is true" would either have the same truth table as each of it's component statements or it would not.  De Morgan's theorem is something that I have read and may have understood.  It is not something that I can use yet.  I know there are some who can just look at it and tell something, but I do not know yet what that something is.  I assume that there may come a time when I will be able to do so.  If that time arrives, it will be because of the patient effort those who wrote the answers below.
There are two ways to approach this question. 


The first one it to start with natural language and with the ordinary expression " if...then" and, from this starting point, to ask ourselves: what do we mean when we utter " if A, then B". For example what do I mean if I do the promise : " If you travel to Mars before the end of this month, I'll buy a car for you.". 


I think you will agree that, the only way for my promise to turn into a false one is the case where (1) the person actually manages to travel to Mars before the end of the month, and (2) I do not buy a car for her. In all other cases ( the 3 remaining ones on the truth table) , I will not have said something false, my promise-statement will be a true one. 


In the second approach, one forgets natural language, starts with the truth table of the " $\rightarrow$ " operator ( without even trying to "read" the symbol in natural language) and considers  this truth table as its definition. By inspecting the truth table, you will see that the proposition $(P{\rightarrow}Q)$ is true on line $1$, $2$, and $4$. That is to say, $(P{\rightarrow}Q)$  is true just in case we are not on line $2$, where $P$ is true and $Q$ is false. Translating " just in case" as a biconditional or as an equivalence, we therefore have 



  $(P{\rightarrow}Q)$  if and only if  $NOT(P\wedge NOT-Q)$


( Read : P implies Q is true iff it is not the case that ( P is true and Q is false). 


*As you see,  in the second approach, there is no real point in asking "why" this is true. This is simply the definition of the " $\rightarrow$" operator.* 
The "construction" of propositional calculus could be explained by the following story:  (1) we have a truth table ( which is one amongst the 16 posssible truth tables for a binary connective). (2) We therefore know that there is such a binary connective with this truth table. (3) Then, we decide to assign to it the symbol " $\rightarrow$ ". (4) And after that, due to the fact we notice that this operator will be useful to define logical implication, we decide to read it as " material implication" or " if...then". 


Note : on the distinction between material implication and logical implication, you may have a look at Seymour Lipschutz, Outline Of Set Theory, Chapter on The Algebra Of Propositions ( at archive.org).
Notice that 
$$P\Longrightarrow Q$$
is the same as
$$\neg P\vee Q$$
which is the same as
$$\neg(P\wedge \neg Q)$$
by de Morgan's law, which is the expression you are asking about.