Show that one cannot infer from the formula ∃xR0(x,c) the formula ∃xR0(c,x)

Question

"Show that one cannot infer from the formula ∃xR0(x,c) the formula ∃xR0(c,x)".

I'm looking for help understanding the difference between the formulas

Answer 1

Since the infering rules are all sound with semantics, if we could infer from $\phi:=\exists x:R(x,c)$ the other one $\psi:=\exists x:R(c,x)$, then in every modell $\mathfrak M$, we should have $\mathfrak M\models\, \phi\Rightarrow\psi$.

I assume that $R$ (what you called 'Ro') is a general binary relation symbol of the language, and $c$ is a general nullary function symbol, that is, a constant. (If $c$ denotes a fixed free variable, the conclusion is going to be the same.)

Now consider a simple model where $\phi\Rightarrow\psi$ fails, that is $\phi$ holds but $\psi$ not. For this, we need at least $2$ elements, one has to be interpreted as the constant $c$. So let $M:=\{a,c\}$, and define $$R^{\mathfrak M}:=\{(a,c)\}\,.$$ This satisfies $\phi$ (there is an $x\in M$, namely $x=a$ such that $\mathfrak M\models R(x,c)$ holds (i.e. $(x,c)\in R^{\mathfrak M}$), but $\psi$ doesn't hold.