12. Invalid deduction: $\neg \forall y(y=x) \vdash \exists x \neg \forall x (x=x)$

Proof (using invalid $\exists I$ rule):

- $\neg \forall y (y=x) \vdash \neg \forall y (y=x)$ by assumption.
- $\neg \forall y (y=x) \vdash \exists x \neg \forall x (x=x)$ by (faulty) $\exists I$ from 1.

Or, in words, we prove that, given not all $y$ is equal to $x$, it is true that there exists $x$ such that $x=x$ does not hold. However, this disagrees with the $=I$ rule, since all $x$ must be equal to $x$.

To reach this method, we must consider the simplest case: one which shows that $x$ does not equal $x$. Since we will be replacing some $y$ with $x$, we use the form$\neg \phi$, where $\phi$ contains $(y=x)$. It is a true statement that $\neg \forall y (y=x)$ and $x$ is not free here, so we choose to use this wff. Then, we simply use the incorrect $\exists I$ and an incorrect statement arises since it claims that there is some $x$ for which $x$ does not equal itself. Note that if we were to use the proper form of $\exists I$ where $y$ must be free in $\phi$, then we get $\neg \forall y (y=x) \vdash \exists x \neg \forall y (y=x)$, which remains true since there is some $x$ for which not all $y$ is equal to $x$.

Comment - Andrew Furash

Looks great!