Can $\neg p \implies p$ be true? How about $p \implies \neg p$?
I was told yes, but it doesn't make sense to me. Any help would be appreciated!
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If $p$ is true, then $\lnot p\to p$ is also true. In fact, $\lnot p\to p$ is equivalent to $p$. – Hanul Jeon Nov 12 '14 at 15:56
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Right, because false $\implies$ true. Thanks! – John Nov 12 '14 at 15:58
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So, in particular, one of $p\implies\lnot p$ or $\lnot p\implies p$ is always true. – Thomas Andrews Nov 12 '14 at 15:58
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Maybe rewrite $\neg p \implies p$ as $\neg\neg p \vee p$. – Najib Idrissi Nov 12 '14 at 16:03
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You can think of the truth of $\neg p\Longrightarrow p$ as itself being like a proof that $\neg p$ cannot possibly be true (since it leads to a contradiction). – Addem Jan 05 '22 at 17:23
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Absolutely. In fact, the general principle $$(\lnot p\implies p)\implies p$$ is often used in mathematical proofs; one shows $\lnot p\implies p$ and from this one concludes $p$.
For example, suppose there are three beans, each either red or blue; we want to show there are at least two beans of the same color. ($p$ is “there are two beans of the same color”.)
Assume the contrary ($\lnot p$), that no two beans have the same color. Then of the first two beans, one must be red and one must be blue; say the red one is bean $B_r$ and the blue one is $B_b$.
Now if the third bean $B_3$ is red then $B_r$ and $B_3$ are the same color, and if it is blue then $B_b$ and $B_3$ are the same color, so there are two beans of the same color. Thus $\lnot p \implies p$, and from this we can conclude $p$, that there must be two beans of the same color.
MJD
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($\lnot p\implies p)\implies p$ however, doesn't answer the question whether $\lnot p\implies p$ is ever true. You either want the implication in the other direction, or an "if and only if." – Thomas Andrews Nov 12 '14 at 16:06
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And a technique where “one shows $¬p⟹p$ and from this one concludes $p$” is useless if one can't ever show $\lnot p\implies p$. So the existence of such a technique, “often used in mathematical proofs”, is strong evidence that $\lnot p\implies p$ is sometimes true. – MJD Nov 12 '14 at 16:10
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In classical logic, yes. But in intuitonistic logic, $p \implies (\lnot p \implies p)$, but $(\lnot p\implies p)\not\implies p$. – MJD Jan 05 '22 at 18:26
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Use the truth table. From the truth table you can see that an implication statement is true if the hypothesis is false. An implication statement can be false only when the hypothesis holds and the conclusion doesn't. So $p\implies \neg p$ holds when $p$ is false and $\neg p\implies p $ holds when $\neg p$ is false.
Praveen
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If $p$ is true (you get $\neg p $ is false) then $p \implies \neg p$ is a false statement, also if $\neg p$ is true (you get $p $ is false) then $\neg p \implies p$ is a false statement. So such statements doesn't hold when the hypothesis is true, and holds if and only if the hypothesis is false. – Praveen Nov 12 '14 at 16:27
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Yes. Because you can imply everything from false, so you need to make sure the first term becomes false.
For: $¬p \rightarrow p$ $$p = true$$ $$(false \rightarrow true)$$ $$\Rightarrow true$$
And for: $p \rightarrow ¬p$ $$p = false$$ $$(false \rightarrow true)$$ $$\Rightarrow true$$
Basti Funck
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