Let $\Sigma$ be a relational language and no relational symbols have an arity greater than $S$. If the second player has a winning strategy for the game $G^S_\infty (A,B)$ and $|A|=|B|\leq S+1$.
I want to show that $A\cong B$.
($S$ is the number of pebble pairs)
First, I defined a function $f: A\rightarrow B$ such that $f$ is a one-to-one correspondence. Then make $f$ keep the structure by defining:
If $(\overline{\alpha_i}:\overline{a_j})\in E^A_k \Rightarrow (\overline{\beta_i}:\overline{f(a_j)})\in E^B_k $
($1\leq i\leq S , 1\leq j\leq |A|$)
By $\alpha_i:a_j$ I mean the pebble $\alpha_i$ is moved in element $a_j$.
Because the number of arities is at most $S$, this is an isomorphism.
I don't know why this is not correct as I have not used the fact that $|A|=|B|\leq S+1$
Is that because in the first-order logic every finite structure can be characterized up to isomorphism by a first-order sentence that needs $|A|+1$ variables?