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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)

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...?

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up vote 7 down vote accepted

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.

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Interestingly enough in 3-valued logic, CpCqp is a tautology even though EANpqCpq is not a tautology. – Doug Spoonwood Jun 8 '14 at 22:42

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".

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Use: $Q\implies P$ is equivalent to $\neg(Q \land \neg P)$

  1. Assume $P$

  2. Assume $Q\land \neg P$

  3. $\neg P$ (Elim $\land$, from 2)

  4. $P \land \neg P$ (Intro $\land$, 1, 3)

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

  6. $Q\implies P$ (Equivalence, from 5)

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

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