Is There a Maximally d-consistent Set that is Not $\omega$-complete? I was going back through Mates 2nd edition Elementary Logic and am having a bit of trouble on one of the exercises; it says,
"Find a counter example to the following assertion: for any set of sentences $\Gamma$, if $\Gamma$ is d-consistent, then there is a set of sentences $\Delta$ that includes $\Gamma$ and is maximal d-consistent and $\omega$-complete."
Now, my thinking is essentially this, if I can find a maximally d-consistent set that is not $\omega$-complete, say, $\Gamma$, then I am done since I could choose $\Delta$ to be $\Gamma$ itself, hence maximally d-consistent but would not be $\omega$-complete (by assumption). But how can I find such a set? Perhaps, my confusion is with what $\omega$-consistency is?
Here are the definitions I have from the book:
d-consistent: a set is d-consistent iff P&-P is not derivable from it.
maximally d-consistent: a set which is d-consistent but is not properly included in any d-consistent set.
$\omega$-complete: a set $A$ is $\omega$-complete iff for every formula $\phi$ and variable $\alpha$, if $(\exists \alpha) \phi$ belongs to $A$, then there is an individual constant $\beta$ such that $\phi \alpha / \beta$ also belongs to $A$.
 A: Mates uses some terminology in non-standard ways. For example:


*

*Mates says "consistent" for what most people call "satisfiable" (has a model).

*Mates says "$d$-consistent" for what most people call "consistent" (doesn't prove a contradiction).

*Mates says "$\omega$-complete" for what most people call "has Henkin witnesses" (for every provable sentence $\exists x\, \varphi(x)$, there is a constant symbol $c$ such that $\varphi(c)$ is provable). 
It also appears that Mates always works in a language with infinitely many constant symbols (Mates says "individual constants") indexed by the natural numbers, $(c_i)_{i\in \omega}$.
Now Mates proves II on p.144: 

For any set of sentences $\Gamma$, if $\Gamma$ is $d$-consistent and all indices of individual constants occurring in the sentences in $\Gamma$ are even, then there is a set of sentences $\Delta$ that includes $\Gamma$ and is maximal $d$-consistent and $\omega$-complete. 

This is exactly the statement you want to find a counterexample to - except it has the additional hypothesis about indices of constants. The point of this is to ensure that there are infinitely many constant symbols which aren't mentioned by sentences in $\Gamma$, so that we have enough constant symbols to serve as Henkin witnesses (to satisfy the definition of $\omega$-completeness). So the way to solve the exercise is to mention all the available constant symbols in the sentences in $\Gamma$ in such a way that there's no available constant symbol to serve as a witness to some existential quantifier. To put it another way, ensure that any model of $\Gamma$ must have an element which is not named by a constant symbol. 
For example: $$\Gamma = \{c_i = c_0\mid i>0\}\cup \{\exists x\, (x\neq c_0)\}.$$
Then $\Gamma$ is consistent (it's easy to find a model), but for any maximal consistent $\Delta\supseteq \Gamma$, we will have that $\Delta$ proves $\exists x\, (x\neq c_0)$, but there is no constant symbol $c_i$ such that $\Delta$ proves $c_i\neq c_0$, since also $c_i = c_0$ is in $\Gamma$ contradicting consistency of $\Delta$. 
Incidentally, a more typical approach to proving completeness, which doesn't have to fiddle with the indexing of the constant symbols, would be to replace II with:

For any set of $L$-sentences $\Gamma$, if $\Gamma$ is consistent, then there is a language $L'$ extending $L$ with infinitely many new constant symbols and a maximal consistent set of $L'$-sentenes $\Delta\supseteq \Gamma$ which has Henkin witnesses. 

That is, just add in new constant symbols, rather than reserving infinitely many from a pre-existing infinite supply. In general, I would recommend learning logic from a more modern textbook. 
