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...?


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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  • $\begingroup$ Interestingly enough in 3-valued logic, CpCqp is a tautology even though EANpqCpq is not a tautology. $\endgroup$ – 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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