Is Robinsonian Semantics a special case of Tarskian Semantics? Like the question says,
Putting all the nitty-gritty aside, Tarksi semantics boils down to an assignment function that maps a sequence of elements in the universe to a set of variables in a term.
Robinson takes another approach:
consider a language $L$ and structure $\scr S$ of $L$ and universe $S$. Instead of including an assignment function, we take an expansion of $L$, $L^+$, which is just $\scr {S}$ $\cup$ $C$ where $C$ includes a constant $c_i$ for each $s_i$ in $S$.
I can't prove this, but I think every Robinsonian term is a Tarksian term with the only difference being: if we consider a Robinsonian term within $\scr S$ it just has a particular assignment-namely- $\bar c$ which is from $C$.
 A: There isn't a substantive difference between the two (and I've never heard the term "Robinson semantics" before - I've seen both approaches subsumed by the term "Tarskian semantics"). Basically, in order to recursively define the satisfaction relation for normal sentences, it turns out that we have to consider some kind of "generalized sentence" along the way due to the behavior of quantifiers. There are two reasonable ways to get such a generalization, namely variable assignments or parameters in the syntax. These do exactly the same thing, but look different on the surface.

Using the terminology in the OP, here's a possibly-overly-detailed description of the situation. Tarskian semantics recursively defines a class $\mathfrak{C}_{Tarski}$ of triples of the form $(\mathcal{A},\varphi,\nu)$ where $\mathcal{A}$ is a structure in some language $L$, $\varphi$ is an $L$-formula, and $\nu$ is an assignment of elements of $\mathfrak{A}$ to the free variables in $\varphi$: we have $(\mathcal{A},\varphi,\nu)\in\mathfrak{C}_{Tarsk}$ iff $\mathcal{A},\nu\models\varphi$ in the Tarskian sense. Similarly, Robinsonian semantics recursively defines a class $\mathfrak{C}_{Robinson}$ of pairs of the form $(\mathcal{A},\hat{\varphi})$ where $\mathcal{A}$ is a structure in some language $L$ and $\hat{\varphi}$ is a sentence in the language $L_\mathcal{A}$ gotten by adding to $L$ a constant symbol for each element of $\mathcal{A}$: we have $(\mathcal{A},\hat{\varphi})\in\mathfrak{C}_{Tarsk}$ iff $\mathcal{A}\models\hat{\varphi}$ in the Robinsonian sense.
The point is that $\mathfrak{C}_{Tarsk}$ and $\mathfrak{C}_{Robinson}$ are really equivalent. For example, fix a structure $\mathcal{A}$ in a language $L$ and an $L_\mathcal{A}$-sentence $\hat{\varphi}$. Let $c_{a_1},...,c_{a_n}$ be the "$\mathcal{A}$-constants" appearing in $\hat{\varphi}$, let $x_1,...,x_n$ be variables not occurring in $\hat{\varphi}$, and let $(\varphi,\nu)$ be the following pair:

*

*$\varphi$ is the $L$-formula gotten by replacing each "$c_{a_i}$" in $\hat{\varphi}$ by "$x_i$," and


*$\nu:\{x_1,...,x_n\}\rightarrow\mathcal{A}: x_i\mapsto a_i$ is the variable assignment "remembering" what we've replaced.
We have $(\mathcal{A},\hat{\varphi})\in\mathfrak{C}_{Robinson}$ iff $(\mathcal{A},\varphi,\nu)\in \mathfrak{C}_{Tarski}$. And we can similarly translate the other way. So (up to annoying variable/constant symbol naming issues) Tarski and Robinson provide exactly the same picture. In particular, we have:

Suppose $\varphi$ is an $L$-sentence and $\mathcal{A}$ is an $L$-structure. Then $$(\mathcal{A},\varphi,\emptyset)\in\mathfrak{C}_{Tarski}\quad\iff\quad(\mathcal{A},\varphi)\in\mathfrak{C}_{Robinson}.$$ (Note that $\emptyset$ is indeed a variable assignment for $\varphi$ since $\varphi$, being a sentence, has no free variables whatsoever.)

