Boolean and witness closure of a complete Henkin theory

If both $\sigma$ and $\mathrm{\neg }\sigma$ belonged to $H$, their assumption proofs and explosion would prove bottom, contrary to consistency. If $\sigma \mathrm{\notin }H$, completeness gives a proof of $\sigma$ or $\mathrm{\neg }\sigma$; the first would put $\sigma$ in $H$ by deductive closure, so the second puts $\mathrm{\neg }\sigma$ in $H$. Conversely $\mathrm{\neg }\sigma \in H$ excludes $\sigma \in H$ by the first argument.

If $\sigma \wedge \tau \in H$, conjunction elimination and closure put both conjuncts in $H$. If both conjuncts belong, the conjunction introduction tautology and two MP applications give their conjunction in $H$. Thus both conjunction directions hold.

If $\mathrm{\exists }x\varphi \in H$, take the single witness constant in F1; MP and closure yield $\varphi [c\mathrm{/}x]\in H$, with $c$ a closed term. Conversely a closed term $t$ has no free variables and is free for $x$ in $\varphi$; applying existential introduction to $\varphi [t\mathrm{/}x]\in H$ and closing gives $\mathrm{\exists }x\varphi \in H$. This proves the existential equivalence even for vacuous $x$. No simultaneous selection of witnesses and no rewriting inside a quantifier is used.