This feels reminiscent of Lindenbaum's lemma ascerting the existence of a maximal consistent set containing a given consistent set. I am wondering if something similar can be done here. First of all, I don't think $\sqsubseteq$ is a partial order unless we look the quotien set ${\cal F}/\sim$ modulo the equivalence $T\sim T'$ defined by $O_{n}(T)=O_{n}(T')$ for all $n$. Given a consistent set $T_{0}$ with equivalence class $[T_{0}]$ we may ask whether ${\cal F}=\{[T]: T\ \mbox{consistent},\, [T_{0}]\sqsubseteq[T]$\} has a maximal element, and hope to apply Zorn's lemma. So we need to check that every chain in ${\cal F}$ has an upper bound in ${\cal F}$. Now if ${\cal C}\subseteq{\cal F}$ is a chain in ${\cal F}$ (we can assume it is not empty), it is tempting to define $T^{*}=\cup\{T: [T]\in{\cal C}\}$ and hope that $[T^{*}]$ is an upper bound (the set $T^{*}$ is not uniquely defined but hopefully $[T^{*}]$ is). Now let us fix a variable name '$x$' and consider the mapping $\tau:{\cal L}-\mbox{Form}\to{\cal L}-\mbox{Form}$ defined by $\tau(\phi)=\forall y_{1}\ldots\forall y_{n}\exists !x\phi$ where it is understood that $\{y_{1},\ldots,y_{n}\}=\mathtt{Fr}(\phi)\setminus{\{x\}}$ (free variables of $\phi$ except $x$). (so $\tau$ is a form of universal closure mapping but not quite). Then it seems to me the equivalence $T\sim T'$ can be expressed as $T\vdash \tau(\phi)\Leftrightarrow T'\vdash\tau(\phi)$ for all $\phi$. Denothing $\mathtt{Th}(T)=\{\phi:T\vdash\phi\}$, the equivalence becomes an equality between inverse images $\tau^{-1}(\mathtt{Th}(T))=\tau^{-1}(\mathtt{Th}(T'))$. Similarly the partial order $[T]\sqsubseteq[T']$ can be expressed as $\tau^{-1}(\mathtt{Th}(T))\subseteq\tau^{-1}(\mathtt{Th}(T'))$. Now coning back to our upper-bound candidate $[T^{*}]$ the key is that since ${\cal C}$ is totally ordered we have $\mathtt{Th}(T^{*})=\cup\{\mathtt{Th}(T): [T]\in{\cal C}\}$ and $\tau^{-1}(\mathtt{Th}(T^{*}))=\cup\{\tau^{-1}(\mathtt{Th}(T)):[T]\in{\cal C}\}$. Furthermore $T^{*}$ is consistent. So I have the feeling this works. Please forgive me if I am completely off the mark.