Question about implication with antecedent $P(x)$ of $x$ that is false for all values of $x$. Suppose $x \in R^+$ and we want to prove the implication $x < 0 \Rightarrow  P(x)$, where $P(x)$ is some statement of $x$.
How should one tackle this situtation ?
Normally one should prove the implication in the case that the antecedent is false and in the case that the antecedent is true, where the false case follows (is true) by definition.
Should one prove the implication in the case $x < 0$ (antecedent is true) even thus $x \in R^+$, or is the case optional (we can just say the truth case will never happen ?).
More generally suppose we don't know whether $P(x)$ can be true for any $x$, and we want to prove $P(x) \Rightarrow Q(x)$.
Is it then O.K to assume $P(x)$ is true if it is false for all values of $x$ ? Should one prove the implication for $P(x)$ assumed to be true in order to prove the implication ?

Since $x\in \mathbb R^+$,  $$x \lt 0 \rightarrow P(x)$$ is always true, because an implication is always true when the antecedent if false.
Remember that the only situation in which an implication $a \rightarrow b$ is false is when $a$ (antecedent) is true AND $b$ (consequent) is false. If needed, refer to the truth-table for $\rightarrow$:

Can you see that the only way an implication $a \rightarrow b$ can be false is if $a$ is true, and $b$ is false?
Since in our case, the antecedent $x \lt 0$ is false, it doesn't matter what $P(x)$ is, the implication as a whole doesn't meet the conditions of falsehood, so is thereby true.
In general: 
$P(x)\rightarrow Q(x)$ (the implication as a whole) is certainly true if $P(x)$ is false for all values of $x$ in the domain, and this holds regardless of the truth value of $Q(x)$
