Section 2.5: Problem 1 Solution »

Section 2.5: Soundness and Completeness Theorems


Every logical axiom is valid.
A1: Exercise 2.4.3(b). A3: Exercise 2.2.3. A4: Exercise 2.2.4. A5: By definition.
A6: Similar to Exercise 2.2.5, but using induction in general.
A2: We prove and use and (Substitution Lemma) (where is substitutable for in ) for terms and formulas, respectively.
A set of wffs is called satisfiable iff there is and such that satisfies (every member of) with .
(Soundness Theorem) If , then .
  • If is satisfiable, then is consistent. This is, in fact, equivalent to the theorem.
  • If or, equivalently, and , then and are logically equivalent.
    • Alphabetic variants are logically equivalent.
(Completeness Theorem; Gödel, 1930) If , then .
  • If is consistent, then is satisfiable. This is, in fact, equivalent to the theorem.
    Let be consistent. First, we expand the language by adding infinitely many constant symbols , and expand by adding wffs for each (in the expanded language) and using constant symbols such that and does not occur in . The resulting set is consistent. Further, we extend to a consistent such that for each , either or (in the uncountable case we use Zorn’s Lemma, but however we do this, turns out to be deductively closed: iff ).
    Second, we define a) a new binary relation (instead of ); b) a structure where is the set of all terms (of the expanded language), , , ; , and c) a function by ( ). Then, iff where is with replaced by .
    Third, we note that is a congruence relation for (an equivalence relation such that and are compatible with ), therefore, the quotient structure is such that a) is a homomorphism of onto where , b) is the equality relation on , and c) iff . The restriction of to the original language satisfies every member of with .
(Compactness Theorem) If , then for some finite , .
  • is satisfiable iff every finite is satisfiable. has a model iff every finite has a model.
  • Disjoint classes can be separated by an class: if for two sets of sentences, , then there is a sentence such that and .
  • For where iff , there is an elementarily equivalent disconnected structure.


(Enumerability Theorem) For a reasonable language (the set of parameters can be effectively enumerated and the relations and are decidable), the set of valid wffs can be effectively enumerated.
  • A finite language is one having only finitely many parameters. A finite language is reasonable.
  • A reasonable language must be countable.
  • If is decidable and the language is reasonable, then and are effectively enumerable.
  • If is decidable, the language is reasonable, and for every sentence , or , then is decidable.