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If you start out with a false premise, then, as far as implication is concerned, you are free to conclude anything. (This corresponds to the fact that, when $P$ is false, the implication $P rightarrow Q$ is true no matter what $Q$ is.)

If you start out with a true premise, then the implication should be true only when the conclusion is also true. (This corresponds to the fact that, when $P$ is true, the truth of the implication is the same as the truth of $Q$.)

Not really as an answer but as an anecdote I'll sketch the following real life situation (in abstract terms to avoid controversy).

A politician $s$ declares, in a menacing voice: "if we would do $P$ then $Q$ will ensue!" where $P$ is something like electing his adversary, or not adopting the Draconic measures he proposes, and $Q$ are catastrophic events like people losing their jobs and the country plunging into a deep crisis. Now suppose $s$ is lucky and manages to avoid $P$, but that then $Q$ happens anyway. Now does this show that $s$ lied? Since $P$ is false and $Q$ is true, we are in the second line of your table, you can read off that $PRightarrow Q$ is deemed true in this case. In fact since $s$ prophesized about a circumstance $P$ that did not happen, later events could not have shown him a liar either way. An this in spite of the fact that by common sense the statement he made was either false (if doing $P$ would actually have prevented $Q$) or irrelevant (if $Q$ would have happened independently of $P$, or depending on other conditions than those of $P$).

You see how smart politicians are? (Of course $s$ can be shown to be a liar if he does not manage to avoid $P$, but then being out of office anyway, $s$ probably won't care much about being proven a liar as well.)

Remember that "implies" is equivalent to "subset of". It works in exactly the same way: "if an element is in the subset (e.g A), it MUST also be in the superset (e.g. B)". By definition, it is impossible that an element is in the subset, but not in the superset. That's the P=1, Q=0; P=>Q = 0 case. In fact, "A ⊆ B" means that a ∈ A implies that a ∈ B. If a is not in subset A then you can't draw any conclusions on whether a is in the superset B. That's how I keep remembering it.

"if P then Q" is equivalent to P=>Q

"P only if Q" is also equivalent to P=>Q (example here)

"Q if P" is same as "if P then Q" and equivalent to P=>Q

However, note (following statement which is not given in the original question): P if Q is equivalent to "if q then p" or Q=>P

In your truth table, look only at the lines where $Pimplies Q$ holds (is $1$), i.e., drop the third line. In the remaining lines, for each line where $P$ holds (i.e., the last one) $Q$ holds as well. Moreover, the column $Pimplies Q$ has $1$s in all lines possible where this property holds (i.e., if we made its unique entry $0$ in the third line a $1$ as well, the property would fail).

Here's a good rule to follow when you're at a university (or working on the job):

"If the fire alarm is going off, you are to go outside"

This can be expressed as P -> Q:

P = "Is the alarm going off?"
Q = "Are you going outside?"

begin{array}{|r|r|r|} text{P} & text{Q} & text{Answer} & text{Have you followed the rule?} \ hline
text{F} & text{F} & text{T} & text{Yes, the rule has been followed} \ hline
text{F} & text{T} & text{T} & text{Yes, the rule has been followed (you can still go outside during a break)} \ hline
text{T} & text{F} & text{F} & text{No, the rule has been broken!} \ hline
text{T} & text{T} & text{T} & text{Yes, the rule has been followed} \ hline
hline
end{array}

Here's your truth table:

$$ begin{align}mathbf P & mathbf Q & mathbf{P implies Q} \ 1 & 1&1 \ 1&0&0 \ 0&1&1 \ 0&0&1end{align}$$

$1$ means true and $0$ means false.

What does logical implication mean? "If $Phi$ then $Psi$" can be written as $mathbf{Phi Rightarrow Psi}$. Albeit, it's much more defined than real life. Remember that Mathematical thinking is different than general thinking.

From the MIT CS Math course

The truth table for implications can be summarized in words as
follows:

An implication is true exactly when the if-part is false or the
then-part is true.

I would like to share my own understanding of this.
I like to think of Implication as a Promise rather than Causality which is the natural tendency when you come across it the first time.

Example:

You have a nice kid and you make him the following promise to him:

If you get an A in your exam, then I will buy you a car.

In this case P is kid gets A in exam and Q is You buy him a car.

Now let's see how this promise holds with various values for P and Q

If P is true (Kid gets A in exam) and Q is true (You bought him car) then your promise has held and $P Rightarrow Q$ is true.

If P is true (Kid gets A in exam) and Q is false (You didn't buy him a car) then your promise didn't hold so $P Rightarrow Q$ is false.

If P is false (Kid didn't get A in exam) and Q is true (You bought him car) then your promise still holds and $P Rightarrow Q$ is true and that's because you only said what will happen if he get's an A, you basically didn't say what will happen if he doesn't which could imply anything. Basically you didn't break your promise and this is the weak property which most people find confusing in implication.

If P is false (Kid didn't get A in exam) and Q is false (you didn't buy him a car) then your promise has also held and $P Rightarrow Q$ is true because you only promised and guaranteed a car if he gets an A.