TITLE: When $\;\text{FALSE}\implies P(x),\;$ is $P(x)\;$ false?
QUESTION [5 upvotes]: Say we know that $P(k) \implies P(k+3)$. Then if we know $P(1)$ is true, we know $P(4), P(8) \dots$ are also true. 
However if we know $P(1)$ is false, does that mean $P(4), P(8) \dots$ are also false?
I'm a little confused on if we can trust an implication which is of the form, false $\implies P(x)$.

REPLY [8 votes]: Here's a simple example to drive the point home in this specific case: let $P(n)$ be the predicate '$n$ is greater than $2$'.  Then if $P(k)$ is true, we have $k+3\gt k\gt 2$, so $P(k+3)$ is true; thus $P(k)\implies P(k+3)$.  What's more, $P(1)$ is manifestly false; $1$ is certainly not greater than $2$.  But in fact, $P(4)$, $P(7)$, $\ldots$ are all true.

REPLY [5 votes]: No, in your example, knowing $P(1)$ is false tells us absolutely nothing about the truth value of $P(n)$, $n> 1$.
Even if we know in advance that the entire implication is true, so that it is true that $P(k) \implies P(k+1)$, and we also learn that $P(k)$ is false, then we still cannot conclude anything about the truth or falsity of $P(k + 1)$, even though the implication is true.
We have defined implication (aka the material conditional), as it is used in logic and in math, such that $$\text{FALSE} \implies P(x)\quad \text{ is always true}.$$
Indeed, by definition: 
$$P \implies Q \quad \text{is  FALSE  if and only if}\;\; P \;\;\text{is true}\;\;{\bf and}\;\; Q \;\;\text{is false}.$$
Recall the truth-table for the implication $P\implies Q$ which gives all possible truth-value assignments to P and Q, and the resultant truth value of $P\implies Q$: