 Solution

In the presence of equality, two syntactically different closed terms can be asserted equal by the Henkin theory, so the raw first-order term domain from part (b) must be replaced by a quotient set. DefineThe equality axioms give reflexivity, symmetry and transitivity, so this is an equivalence relation. Their substitution consequences make it a congruence relation on a first-order structure: replacing an input by an equivalent first-order term preserves equivalence of function values and preserves membership of atomic formulas in $H$.

Take domain $\{[t]:t\text{ is a closed term}\}$. Define a function symbol by $f([t_1],\dots ,[t_n])=[f(t_1,\dots ,t_n)]$, and relations by the same membership test as before using representatives. The congruence relation on a first-order structure makes these definitions independent of representatives. Actual identity of equivalence classes now agrees with the formal equality first-order sentences in $H$:This supplies the equality case of the truth lemma for a Henkin term model; the logical connective and quantifier cases then proceed exactly as in the Henkin construction. Thus the term model built from the quotient set is a genuine first-order model with equality interpreted as identity.