Let $T$ be a set of closed literals of signature $L$. Show that a) is equivalent to b). a) Some $L$-structure is a model of $T$. b) If $\neg \phi$ is a negated atomic sentence in $T$, then $\phi$ is not in the $=$-closure of the set of atomic sentences in $T$.
I don't see how this isn't plainly false. We have a theorem:
For any signature $L$, if $T$ is a set of atomic sentences of $L$ then there is an $L$-structure $A$ such that: a) $A \models T$. b) every statement of $A$ is of the form $t^A$ for some closed term $t$ of $L$. c) if $B$ is an $L$-structure and $B \models T$ then there is a unique homomorphism $f\colon A\to B$
In light of this theorem, why isn't the problem false? The theorem says that for ANY set of atomic sentences there is an $L$ structure that models $T$. So why isn't a true by default? What is the issue here?
Edit: Is it because a closed literal (might be) a negation of an atomic sentence so the theorem doesn't apply?
Hint
In predicate logic, an atomic formula is a formula $P_nt_1\ldots t_n$, where $P_n$ is an $n$-place predicate symbol and $t_1,\ldots, t_n$ are terms.
A literal is an atomic formula or its negation.
Thus, a closed literal is an atomic formula or its negation with all terms $t_i$ closed.
Examples from the language of set theory :
The two statements of the question do not contradict each others exactly because in one case (the atomic sentences of the theorem) there are no formulae with a negation in front; thus, loosely speaking, no contradiction can arise.
As you say, in the case of the exercise, the theorem does not apply because the set of closed literals can have formulae like : $\lnot P_nt_1\ldots t_n$.