On logical connector implies It may appear at first glance that this question had been asked over and over here. But I feel that the question that is in my mind is slightly different from what has already been asked. Here it is:

What would have happened if $P \implies Q$ was taken false when $P$ was false and $Q$ was true?

I read several answers for the question: "Why false implies True is false?". I could gather some information from the answers. One is that irrespective of the truth value of $Q$, if $P$ is false, we treat $P \implies Q$ to be true vacuously. But this didn't answer my question, what would have happened if I had taken it to be false?
Another answer, which though addressed my question, I wasn't much satisfied with. This was that answer: If we had taken false implies true to be false, then truth tables of $\implies$ and $\leftrightarrow$  would have been one and the same.
Any answer in the direction of consequence of false implies true false is highly appreciated. I mean, would there have been any logical fallacy, contradiction or paradox of so kind if I had assumed false implies true to be false?
 A: This is probably more of a comment on @Dan's answer, but I feel I have something important to say so I'm making it an answer.
The first thing to be clear about is that the truth table for implication is a definition.  In other words, it is not a statement about the real world, but a statement about how we intend to use language.  Or in other words again: it is not something which has to be true, but something which we mean to be convenient, something that matches our intuitive conception of "if... then".
So, what features characterise our intuitive understanding of "if... then"?  Probably the most important is

*

*given the statement "if $p$ then $q$" and the statement $p$, we may deduce $q$,

sometimes known as modus ponens.  Another is modus tollens:

*

*given the statement "if $p$ then $q$" and the statement that $q$ is false, we may deduce that $p$ is false.

We would also want it to be possible in some cases for "if $p$ then $q$" to be false - not much point discussing it if it is always true.
The only truth table for $p\to q$ which matches our intuitive understanding (in the above sense) is the accepted one.  See Dan's answer for the details.

Addendum.  After a bit of thought, here is another way of doing it which may be more intuitively appealing.  Our basic intuitive properties of implication are:

*

*any statement implies itself (so $p\to p$ is always true);

*modus ponens is valid (from $p\to q$ and $p$ we can deduce $q$);

*the converse error is not valid (from $p\to q$ and $q$ we cannot deduce $p$).

A: Drawing out a truth table makes it easy to see this.
$$\begin{array}{ccc} P & \implies & Q \\ \hline 
F & T & F \\ F & F & T 
\\T & F & F \\ T & T & T \end{array}$$
Nothing more and nothing less than the equivalence operator.
A: Let $f_{abcd}(x, y)$ denote a function of two Boolean variables defined by the truth table:
 x │ y │ f
───┼───┼───
 0 │ 0 │ a
 0 │ 1 │ b
 1 │ 0 │ c
 1 │ 1 │ d

There are 16 possible combinations of $a$, $b$, $c$, and $d$, defining 16 possible functions.  Let's try to find one such that $f_{abcd}(x, y)$ is as similar as possible to the statement “If $x$, then $y$” under classical rules of logical inference.
Modus ponens
A syllogism of the form:

*

*(premise) If P, then Q.

*(premise) P.

*(conclusion) Therefore, Q.

Or, in symbolic notation, $((P \rightarrow Q) \land P) \rightarrow Q$.
(The distinction between $\implies$ and $\rightarrow$ doesn't matter for my purposes here, so I'll use $\rightarrow$ throughout just for aesthetic reasons.)
Or, expressing $\rightarrow$ as a function $f$, then $f(f(P, Q )\land P), Q)$.
Of the 16 possible $f$ functions, four of them satisfy the modus ponens law for all possible combinations of P and Q: $f_{1100}$, $f_{1101}$, $f_{1110}$, and $f_{1111}$.  That is, $a$ and $b$ must both be 1, but $c$ and $d$ may be either 0 or 1.
$f_{1101}$, the conventional $\rightarrow$ operator, satisfies modus ponens.  But $f_{1001}$ (aka $\leftrightarrow$) does not satisfy it.  Specifically, it fails when $P \oplus Q$.
Modus tollens

*

*(premise) If P, then Q.

*(premise) Not Q.

*(conclusion) Therefore, not P.

In symbolic notation, $((P \rightarrow Q) \land \lnot Q) \rightarrow \lnot P$.
In function notation, $f(f(P, Q) \land \lnot Q, \lnot P)$.
This rule is satisfied only by $f_{1101}$ (the conventional $\rightarrow$ operator) and $f_{1111}$ (the constant True function).
Existence of false conditionals
$\exists P, Q : \lnot(P \rightarrow Q)$.
This rules out $f_{1111}$, but the other 15 functions all qualify.
Putting it all together
Of the 16 possible ways to define $\rightarrow$ as a function of two Boolean parameters, only one simultaneously satisfies modus ponens and modus tollens while allowing false conditional statements to exist.
 x │ y │ x→y
───┼───┼─────
 0 │ 0 │  1
 0 │ 1 │  1
 1 │ 0 │  0
 1 │ 1 │  1

Changing the second row of this table (effectively turning $\rightarrow$ into $\leftrightarrow$) would break both modus ponens and modus tollens.
A: You're not going to get anything like a logical paradox or anything because the way you are considering defining it is not self-contradictory. In fact, as @Prime Mover and you both point out, this would make it equivalent to $\Leftrightarrow$.
Nevertheless, this would lead to what would probably be considered an unsatisfactory operator. For example, "$x < 5 \Rightarrow x < 10$" would be false which is not what we want from an "if ... then ..." statement.
A: 
What would have happened if $P \implies Q$ was taken false when $P$ was false and $Q$ was true?

To avoid inconsistencies, at the very least, you would have to invalidate at least one step in the following proof that $\neg P \implies [P \implies Q]:\\$

(Screenshot from my proof checker)
You might, for example, have to disallow or somehow restrict proof by contradiction (line 5), or elimination of '$\neg\neg$' (line 6).
