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?
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?
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.
$a \implies b$ 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.