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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?

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  • $\begingroup$ The statement $A\implies\neg A$ is satisfiable when $A$ is false. $\endgroup$ – Robert Wolfe Mar 6 '14 at 18:29
  • $\begingroup$ Truth table shows that the sentence $A\implies B$ is true whenever $A$ is false. Replace $B$ by $\bar{A}$. You are right, the sentence is not always false. $\endgroup$ – André Nicolas Mar 6 '14 at 18:29
  • $\begingroup$ .. When $A=false$, $A \to \lnot A$ is $true$ $\endgroup$ – user76568 Mar 6 '14 at 18:29
  • $\begingroup$ Note that the contraositive manipulation does not show that the sentence is sometimes true. $\endgroup$ – André Nicolas Mar 6 '14 at 18:33
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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).

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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 ;-)

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  • $\begingroup$ When you says "$A \rightarrow B$" you are not asserting $A$ : you are asserting the conditional "if $A$, then $B$" ... $\endgroup$ – Mauro ALLEGRANZA Mar 6 '14 at 18:41
  • $\begingroup$ @MauroALLEGRANZA I know that. I'm just saying this could be a source for confusion. $\endgroup$ – user76568 Mar 6 '14 at 18:45

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