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Consider the following axiom:

$\lnot a \implies a$

Intuitively, this seems like a contradiction. But all implications hold if the LHS is false. Does this mean that:

$a$

is a valid conclusion? Or is there a contradiction in the axiom?

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    If you take $\neg a \to a$ as an *axiom*, you take it as true. It is possible for an axiom to be contradictory, IIRC, but then *all* formulae are valid conclusions. (Your's isn't and $a$ is$a$valid conclusion).2012-06-05

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This is a tautology, and since there is only one free variable, it is easy to check.

However, you can simplify the expression symbolically before. $\lnot a \implies a$ is logically equivalent to $\lnot(\lnot a)\lor a$, which is $a$. Hence your formula is just $a \implies a$, which is more obviously a tautology.

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    @HenningMakholm I agree with you, and I am myself$a$constructivist militant. This was$a$classical “obviously” ;)2012-06-05
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$a \implies b$ is equivalent to $\neg a \vee b$ so $\neg x \implies x$ implies $\neg \neg x \vee x$ which, unless we are working with constructive logic, implies $x$. This means that $(\neg x \implies x) \implies x$ is an axiom. $\neg x \implies x$, however, is not. Just take $x = 0$ and see that $1 \implies 0$ is false.