Law of excluded middle says that for any proposition ($A$) either it is true or it's negation ($\bar{A}$) is true: $A\veebar\bar{A}$.
When I was taught math logic, this was given as an axiom, but I thought of a proof of this law. My proof goes like this:
1) Let's assign letter $L$ to mean the law of excluded middle. If assume that it is correct, it instantly follows that $L$ is correct. Let's instead assume that it is wrong: $\bar{L}$.
2) If we use law of excluded middle, we would get that $L\veebar\bar{L}$. But we assumed that this is wrong. And the law could be wrong in two ways – either both $L$ and $\bar{L}$ could be wrong or both are correct.
3) We have assumed that $\bar{L}$ is true, so it can't be that they are wrong, thus both must be correct. We have shown that $L$ is true regardless of our assumptions, so it is proven.
Is this proof alright or it logically fails somewhere?
Does this interfere with Gödel's completeness stuff in any way?
Best Answer
You have a problem in that assuming $L$ to be false does not mean that $L\veebar\bar{L}$ is false. It just means that $A\veebar\bar{A}$ is false for some statement $A$, which may not be $L$.