When is implication true? If we have $p\implies q$, then the only case the logical value of this implication is false is when $p$ is true, but $q$ false.
So suppose I have a broken soda machine - it will never give me any can of coke, no matter if I insert some coins in it or not.
Let $p$ be 'I insert a coin', and $q$ - 'I get a can of coke'.
So even though the logical value of $p \implies q$ is true (when $p$ and $q$ are false), it doesn't mean the implication itself is true, right? As I said, $p \implies q$ has a logical value $1$, but implication is true when it matches the truth table of implication. And in this case, it won't, because $p \implies q$ is false for true $p$ (the machine doesn't work).
That's why I think it's not right to say the implication is true based on only one row of the truth table. Does it make sense?


When is implication true?

$p\implies q$ is true when both $p$ and $q$ are true, or when $p$ is false. Assuming your problem is with the latter.
Theorem:
$\neg p \implies [p \implies q]$
Proof:
Suppose $\neg p$.
Suppose $p$.
Suppose to the contrary that $\neg q$.
We have the contradiction $p\land \neg p.$
Therefore $\neg\neg q$ by contradiction.
Remove $\neg\neg$ to obtain $q$.
Therefore $p\implies q$.
Therefore, as required, $\neg p \implies [p \implies q]$.
Not sure why this is thought to not hold in natural language, but it is widely believed. Anything whether true or false will follow from a falsehood. What is wrong with: If pigs could fly, I would be the King of France. (Assuming pigs can't actually fly. See If Pigs Could Fly at my math blog.)
