Empty assignment (model theory) Reading through “logic of mathematics” (page 42) I came across the following example:
$$\mathbb{A}\models \forall x(x\le x)[\emptyset] \text{ iff for each } a\in A, \mathbb{A}\models (x\le x) [\emptyset (a/x)] \text{ iff for each } a\in A, x[a/x] \le^A x[a/x]$$
I’m not understanding the second “iff”, in particular how did the empty assignment disappear? I was thinking it ought to be something like $\text{ iff for each } a\in A, x[\emptyset(a/x)] \le^A x[\emptyset(a/x)]$ and I’m basing this off the definition given on page 40 which says that for atomic formulas you have the following:
$$\mathbb{A}\models r(t_1,\ldots, t_n)[p] \text{ iff } 
r^A(t_1[p],\ldots, t_n[p])$$
where is an assignment.
 A: The assignments $[\emptyset(a/x)]$ and $[a/x]$ are the same thing: adding $\emptyset$ to $a/x$ doesn't result in anything different. Remember that a (incomplete) variable assignment for a formula $\varphi$ is a (partial) map from the free variables of $\varphi$ to elements of the structure. Concatenation (a la "$\emptyset(a/x)$") then corresponds to taking the union of the indicated functions. For example, "$t[(a/x)(b/y)]$" means "take the term $t$, replace each free $x$ by $a$, and replace each free $y$ by $b$," and we have $$t[(a/x)(b/y)]=(t[a/x])[b/y]=(t[b/y])[a/x]$$ (we can do the replacements all at once, or first $x$ then $y$, or first $y$ then $x$).
Now what about "$t[\emptyset(a/x)]$" (for $t$ some term)? Well, just remember that $\emptyset\cup A=A$ for all $A$. The expression "$t[\emptyset(a/x)]$" means, intuitively, "Take $t$, do all the substitutions that $\emptyset$ tells you to do, and also replace every free $x$ by $a$." Since $\emptyset$ is, well, empty, it doesn't tell you to do any substitutions, so you're just looking at replacing every free $x$ by $a$ - that is, $t[a/x]$.
