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$A \Longrightarrow \tilde A$ is always false? So here's what I did:

I did the contrapositive $\tilde(\tilde A) = A \Longrightarrow \tilde A$ So by contrapositive, I also get $A$ implies $\tilde A$. I also did a truth table. But I don't get a tautology or contradiction. So it can't always be false. Correct?

Don Larynx
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4 Answers4

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No, it isn't always false. Suppose $A$ is false, so $\tilde A$ is true. Then "$A\implies\tilde A$" is true; any implication with a false protasis is always a truth ("if the moon is made of blue cheese, then pigs can fly" is a true statement).

MPW
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No, $A \rightarrow \lnot A$ is not always false.

It is simply :

$\lnot A$.

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$A \rightarrow \neg A$ is equivalent to $\neg A \vee \neg A$ which is equivalent to $\neg A$

$(A \rightarrow \neg A) \Leftrightarrow \neg A$

Another way to see it is that the original formula is false if $\neg A$ is true but $A$ isn't. This is when $A$ is false.

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It can seem confusing (It did to me at the time), because you might mistakenly $assert$ $A$ when you look at $A \to \lnot A$ without noticing ;-)

user76568
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