Could somebody please explain why this statement is false?
If $\Delta \vDash \exists x.p(x)$, then $\Delta \vDash p(\tau)$ for some ground term $\tau$.
It is a problem from the MOOC "Introduction to Logic" by Stanford on Coursera.
EDIT: I thought that this is just the definition of the $\exists$ quantifier, i.e. $\Delta \vDash \exists x.p(x)$, is equivalent to $\Delta \vDash p(\tau)$ for some $\tau$. I'm looking for an answer that would give an intuitive explanation or an example, where this is false.
EDIT: In the same chapter a very similar looking statement is given as true,
If $\Delta \vDash \forall x.p(x)$, then $\Delta \vDash p(\tau)$ for every ground term $\tau$.
Best Answer
Structures can have elements which are not named by ground terms. In fact, if there are no constant symbols in the language, then there are no ground terms at all!
For a very simple example, consider the empty language (no constant, function, or relation symbols). Then $\exists x\,(x=x)\models \exists x\, (x=x)$, but there is no ground term $\tau$ such that $\exists x\,(x=x)\models \tau=\tau$, simply because there are no ground terms.
For a less trivial example, let $T$ be the theory of groups in the language $\{\cdot,^{-1},e\}$, where $e$ is the identity element. An example of a ground term in this language is $(e\cdot e)\cdot (e^{-1}\cdot (e\cdot e))$. Note that if $G$ is a group (i.e. a model of $T$), then every ground term evaluates to the identity element in $G$. So if $G$ is non-trivial (i.e. if $G$ contains non-identity elements), then $G$ contains elements which are not named by ground terms.
Let $T' = T\cup \{\lnot \forall x\forall y\,(x\cdot y = y\cdot x)\}$. Then $T'$ is the theory of non-abelian groups. We have $$T'\models \exists x\, x\neq e$$ since any model of $T'$ is a non-trivial group. But there is no ground term $\tau$ such that $T'\models \tau\neq e$.
Addressing your edit:
$\Delta\models \exists x\, \varphi(x)$ means: For every model $M\models \Delta$, there is an element $m\in M$ such that $M\models \varphi(m)$. There is no reason why the element $m$ should be the interpretation of a ground term.
$\Delta\models \forall x\, \varphi(x)$ means: For every model $M\models \Delta$, and for every element $m\in M$, we have $M\models \varphi(m)$. In particular, for every ground term $\tau$, if $\tau^M$ is the interpretation of $\tau$ in $M$, we have $M\models \varphi(\tau^M)$. Since this is true for any model $M$, we have $\Delta\models \varphi(\tau)$.