Logical consequence and partial orders First order logic (FOL) has an algebraic formulation in cylindric algebras, which are Boolean algebras supplemented with cylindrification operators. It is often said that the partial order $\leq$ on a Boolean algebra can represent (amongst other things) the logical consequence relation in classical logic. It has also been discussed here: Entailment relations that are not partial orders .
  I don't understand why the consequence relation is a partial order. My confusion regards $\textbf{(Antisymmetry)}$: 
$$(i) \hspace{0.3cm} If \hspace{0.3cm} a ≤ b \hspace{0.3cm} and \hspace{0.3cm} b ≤ a \hspace{0.3cm} then \hspace{0.3cm} a = b \hspace{0.5cm} \textbf{(Antisymmetry)}$$
Suppose $\thinspace$ ≤ $\thinspace$ is the derivability relation of FOL, $\thinspace \vdash$ $\thinspace$. By $\textbf{(Antisymmetry)}$ that would mean we have:
$$(1) \hspace{0.3cm} If \hspace{0.3cm} a \hspace{0.3cm} \vdash \hspace{0.3cm} b \hspace{0.3cm} and \hspace{0.3cm} b \hspace{0.3cm} \vdash \hspace{0.3cm} a \hspace{0.3cm} then \hspace{0.3cm} a = b \hspace{0.5cm} \textbf{(Antisymmetry $\thinspace: \thinspace \vdash$})$$ 
But this doesn't seem correct in the case of FOL if we set $a := \phi$ and $b := \phi \lor \phi$ $\hspace{0.3cm}$(as in $(2)$ below), at least if $'='$ is to indicate syntactic identity, since $\phi \neq \phi \lor \phi$
:
$$(2) \hspace{0.3cm} If \hspace{0.3cm} \phi \hspace{0.3cm} \vdash \hspace{0.3cm} \phi \hspace{0.3cm} \lor \hspace{0.3cm} \phi \hspace{0.3cm} and \hspace{0.3cm} \phi \hspace{0.3cm} \lor \hspace{0.3cm} \phi \hspace{0.3cm} \vdash \hspace{0.3cm} \phi \hspace{0.3cm} then \hspace{0.3cm} \phi \hspace{0.3cm} = \hspace{0.3cm} \phi \hspace{0.3cm} \lor \hspace{0.3cm} \phi \hspace{0.5cm}$$ 
In which case, what is the intended interpretation of $\thinspace$ '=' $\thinspace$ in this context, if not syntactic identity?
Furthermore, suppose that $\thinspace$ ≤ $\thinspace$ is the semantic consequence relation of FOL, $\thinspace \models$ $\thinspace$. By $\textbf{(Antisymmetry)}$ that would mean we have:
$$(1) \hspace{0.3cm} If \hspace{0.3cm} a \hspace{0.3cm} \models \hspace{0.3cm} b \hspace{0.3cm} and \hspace{0.3cm} b \hspace{0.3cm} \models\hspace{0.3cm} a \hspace{0.3cm} then \hspace{0.3cm} a = b \hspace{0.5cm} \textbf{(Antisymmetry $\thinspace: \thinspace \models$})$$ 
But this doesn't seem correct in the case of FOL if we set $a := \phi$ and $b := \phi \lor \phi$ $\hspace{0.3cm}$(as in $(3)$ below), at least if $'='$ is to indicate syntactic identity, since $\phi \neq \phi \lor \phi$:
$$(3) \hspace{0.3cm} If \hspace{0.3cm} \phi \hspace{0.3cm} \models \hspace{0.3cm} \phi \hspace{0.3cm} \lor \hspace{0.3cm} \phi \hspace{0.3cm} and \hspace{0.3cm} \phi \hspace{0.3cm} \lor \hspace{0.3cm} \phi \hspace{0.3cm} \models \hspace{0.3cm} \phi \hspace{0.3cm} then \hspace{0.3cm} \phi \hspace{0.3cm} = \hspace{0.3cm} \phi \hspace{0.3cm} \lor \hspace{0.3cm} \phi \hspace{0.5cm}$$ 
In which case, what is the intended interpretation of $\thinspace$ '=' $\thinspace$ in this context, if not syntactic identity? Is it identity of truth value?
A further question: in FOL = is usually only defined between terms, and not between formulas. Is therefore $\thinspace$ '=' $\thinspace$ in the definition of $\textbf{(Antisymmetry)}$ doing double duty for both $\thinspace$ '=' $\thinspace$ between terms and $\thinspace$ '$\equiv$' $\thinspace$ between formulas? How can this be the case? Is the identity '=' in the definition of $\textbf{(Antisymmetry)}$ representing some aspect of the metalanguage of FOL?
 A: I guess you are considering "$=$" as a syntactic equivalence, which I think in this context is wrong. More than that, even in the original context of partially ordered sets I don't think it is right, and we can use it only because sets are in general simple structures.
Consider a set $\{0,1,2,\ldots\}$ of numbers and set $\{x_0,x_1,\ldots\}$ of variables. Now if we use syntactic equivalence under valuation $f(x_i) = 0$, $x_i \leq x_j$ and $x_j \leq x_i$ imply together $x_i = x_j$ for any $i$ and $j$, but two different variables are never syntactically equivalent. Still, $f(x_i) = 0 = f(x_j)$, which is fine.
Thus, we could say $\alpha \leq_f \beta$ if in valuation $f$ object $\alpha$ evaluating to true implies $\beta$ evaluates to true as well. However, we should then use "$=_f$" which takes the valuation into account too.
We could define $\alpha \leq \beta$ if for all valuations $f$ we have $\alpha \leq_f \beta$, yet then $\alpha =\beta$ should be defined as $\alpha =_f\beta$ being true for all $f$.
The above were only examples. If the structures you are considering are complex, the equivalence you are using should reflect that as well. In fact, what we would like to do, is to define $\leq$ on some equivalence classes. E.g., if $\phi \leq \psi$ means $\phi \vdash \psi$, then together with $\psi \leq \phi$ they imply that $\psi$ and $\phi$ are provably equivalent.
I hope this helps $\ddot\smile$
