P entails Q implies P I have been looking at the following:
P entails Q implies P

And developed the proof as follows:
1. P
2. Q (Start of new subproof)
2.1 P (By 1)
3 Q implies P by INTRODUCTION OF IMPLICATION 2, 2.1

However, while it makes sense in terms of logic, I can't get it's general meaning.
To me, this is like saying: "I have P. Assuming I have Q, I still have P. So Q must imply P."
Or: "It's raining. Assuming I don't have an umbrella, it's still raining. So the fact I don't have an umbrella implies it's raining."
Is this proof simply stating that whatever assumption, the base assumptions will still hold, so it's trivially true, or...?
 A: One nice way to make sense of what is happening: We have that $P$. 
Assume we have $Q$, and reassert what we know we have: $P$.
Then we conclude that "If we have $Q$, then we have $P$, which we symbolize by $\;Q\rightarrow P$.
It so happens that in this case, we have $P$ even when we don't have $Q$, but that doesn't matter, since any implication with a true consequent (in this case $P$) holds. 

Note also that from the premise $P$, we can derive $\lnot Q \lor P$ by $\lor$ introduction. And $\lnot Q\lor P \equiv Q\rightarrow P$, and indeed that equivalence can be derived if needed.
A: No, it's just that it's only valid to say that "$Q \Rightarrow P$" is  false if $Q$ is true and $P$ is false. Suppose your friend calls you up and tells you "If I manage to find a cab, I'll be there in five minutes". And he does show up in five minutes, but not in a cab. He hitched a ride. Would you say your friend lied? No, he just said that if he gets a cab, he'd reach in five minutes. He didn't say "otherwise I won't".
A: Use: $Q\implies P$ is equivalent to $\neg(Q \land \neg P)$


*

*Assume $P$

*Assume $Q\land \neg P$

*$\neg P$ (Elim $\land$, from 2) 

*$P \land \neg P$ (Intro $\land$, 1, 3)

*$\neg (Q \land \neg P)$ (Conclusion, from 2, 4)

*$Q\implies P$ (Equivalence, from 5)

*$P\implies (Q\implies P)$ (Conclusion, from 1, 6)
