For
a) $\phi(x)$ is $(\forall y(y=1+1 \implies x=y))$
b) $\phi(x)$ is $(\forall x(x=1+1 \implies x=y))$
The answer is supposed to be a) but I don't know why. I guess I don't fully understand the notation $M \models \phi(x/2)$ and am therefore not really sure what I am meant to be doing. Any help would be appreciated, thanks!
$\phi(x/2)$ means the formula obtained fron $\phi(x)$ (with $x$ free) when the number $2$ is assigned as value to the variable $x$.
Thus, the question :
often written as : $M \vDash \phi(x)[2]$, ask if the formula $\phi(x)$ is true in $M$ when the value $2$ is assigned to the free variable $x$.
Now, for
a) when $\phi(x)$ is $∀y(y=1+1 \rightarrow x=y)$
we have that :
is $∀y(y=1+1 \rightarrow 2=y)$, which is true in $\mathbb N$.
For
b) when $\phi(x)$ is $∀x(x=1+1 \rightarrow x=y)$
instead, $x$ is not free in $\phi(x)$ and thus we cannot assign $2$ s value to it.
Thus, $\phi(x)[2]$ is simply $\phi(x)$ which is not true : it is enough to assign $1$ to $y$ and we get : $∀x(x=1+1 \rightarrow x=1)$.