6.1. LetLbe a language in L1, letφbe an interpretation ofL, and letTφLbe the set ofφ-true formulas. In§3 it was shown that the setTφLisG¨odelian: it is complete, does not contain a contradiction, is closed with respect to deduction, and contains all the logical axioms Ax L. We say that a set of formulas E
in L is consistent if the set of formulas deducible from E does not contain a contradiction, i.e., if there is no P such that E P and E ¬P; otherwise, we say that E is inconsistent. The basic purpose of this section is to prove the following converse of the result in §3:
6.2. Theorem(G¨odel)
(a) Any G¨odelian set T is the set ofφ-true formulas TφLfor a suitable inter- pretation ofL in some setM having cardinality card (alphabet of L) + ℵ0.(Here and below we always mean the cardinality of the alphabet without the variables.)
(b) Any set of formulasE which containsAxLand is consistent can be imbed- ded in a Godelian set.
The model M which is constructed in the proof consists of expressions in some extension of the alphabet ofL, and thus has a somewhat artificial charac- ter. In the next section we show that, if we are given some natural interpretation (M, φ) ofL, then we can find a submodel having cardinalitycard (alphabet ofL) +ℵ0.
6.3Corollary. (Deducibility criterion).LetE ⊃AxL.
(a) A formula P is deducible from E if and only if eitherE is inconsistent, or P isφ-true for all modelsφof the setE having cardinalitycard(alphabet of L)+ℵ0.
(b) A formula P is independent ofE if and only if bothE ∪ {P} andE ∪ {¬P} are consistent; by Theorem 6.2, this is true if and only if E ∪ {P} and E ∪ {¬P}have models.
In what follows we shall often omit the verification that various formal deductions exist. If the reader wants to fill in such a verification, this can almost always be done more easily using deducibility criterion 6.3 than directly.
Proof of the corollary
(a) If E is inconsistent, then any formula can be deduced from E (Proposition 4.2). Suppose E is consistent and P is φ-true for all models of E. Let ¯P = ∀x1· · · ∀xnP be the “closure” of P. To prove that E P. we consider two cases.
(a1) E ∪ {¬P¯} is inconsistent. Then E ∪ {¬P¯} P, so that, by the¯ Deduction lemma, E ¬P¯ ⇒ P¯. The tautology (¬P¯ ⇒ P¯) ⇒ P¯ and MP giveE P, and then the axiom of specialization and MP give¯ E P.
(a2)E ∪ {¬P¯} is consistent. Then, by Theorem 6.2, the set E ∪ {¬P¯} has a model. In this modelE is true andP is false, so that this case is impossible.
(b) Suppose that P is independent of E, i.e., neither P nor ¬P is deducible.
Then, by part (a), there exists a model ofE in whichP is true and a model of
E in whichP is false. The converse is obvious.
We now proceed to the proof of G¨odel’s completeness theorem.
6.4. Definition. LetE be a set of formulas in a languageL. The alphabet of L is said to be sufficient forE if, for each closed formula ¬∀xP(x) in E there exists a constantcP (depending onP) such that the formula
RP:¬∀x P(x)⇒ ¬P(cP) belongs toE.
The intuitive meaning ofRP is; “If not allxhave the propertyP, then some concrete objectcP can be found that does not have this property.” We say that the alphabet (rather than E) is “sufficient” or “insufficient” because ifE does not contain enough formulas of the type RP, we can simply add all theRp to E, while if there are not enough constantscP, we then have to add them to the alphabet of the language.
The plan for proving Theorem 6.2 is as follows. We first prove the funda- mental lemma:
6.5.Fundamental Lemma.If a set of formulasEin a language L is consistent and complete and contains Ax L, and if the alphabet of L is sufficient for E, then E has a model with cardinality card(alphabet ofL) + ℵ0.
The next two lemmas allow us to embed any consistentEin a complete set, or in one for which the alphabet is sufficient.
6.6.Lemma.If E is consistent and containsAxL,then there exists a consis- tent and complete set of formulas E ⊃ E.
6.7. Lemma.If E is consistent and containsAx L,then there exist:
(a) a languageL whose alphabet is obtained from the alphabet ofL by adding a set of new constants having cardinality card(alphabet ofL) +ℵ0. (b) a set of formulas E in L that is consistent, contains E and Ax L, and
has the property that the alphabet ofL is sufficient forE.
However, these constructions get in each other’s way. If we complete a setE for which the alphabet is sufficient, we might obtain a set with an insufficient alphabet; if we add new constants, we increase the overall supply of formulas in the language, and thereby lose the completeness of E. Hence, we have to alternate the constructions in 6.6 and 6.7 a countable number of times in order to prove our last lemma:
6.8. Lemma.If E ⊃AxLis consistent, then there exist:
(a) a languageL(∞)whose alphabet is obtained from the alphabet ofLby adding a set of new constants having cardinality card(alphabet ofL) +ℵ0. (b) a set of formulas E(∞) inL(∞) that is complete and consistent, containsE
andAx L(∞), and has the property that the alphabet of L(∞) is sufficient forE(∞).
After Lemma 6.8 is proved, Theorem 6.2 is obtained from the fundamental lemma applied toE(∞)if we restrict the resulting model to LandE.
We now prove the lemmas. The fundamental lemma is proved in 6.9, and Lemmas 6.5, 6.6, and 6.7 are proved in Sections 6.10, 6.11, and 6.12, respectively.
6.9.Proof of the Fundamental Lemma. We begin by explicitly construct- ing the interpretationφofLthat will be our model forE.
(a) By aconstant termwe mean a term inLthat does not contain any symbols for variables. We letM ={¯t|t is a constant term} be a “second copy” of the set of constant terms, and we define theprimary mappingsof the interpretation φofLin M as follows:
φ(c) = ¯c (for any constant c);
φ(f)(¯t1, . . . ,¯tr) =f(t1, . . . , tr) (for each operation symbol fof
degreerand all constant termst1, . . . , tr);
¯t1, . . .¯tr ∈φ(p) if and only ifp(t1, . . . , tr)∈ E (for each relationpof degreer and all constant termst1, . . . , tr).
We now prove the following claim:
(b) Claim. Let P be a closed formula. Then |P|φ = 1 if and only if P ∈ E. (This claim implies thatφis a model forE. In fact, ifP ∈ Eis not closed, then its closure ∀x1· · · ∀xnP is deducible from E using Gen, and hence, since E is complete and consistent, ∀x1· · · ∀xnP ∈ E. By the claim, |∀x1· · · ∀xnP|φ= 1, so that|P|φ= 1.)
Proof of the claim. We use induction on the total number of quantifiers and connectives in P. We shall write|P|instead of|P|φ.
(b1)P is an atomic formula p(t1, . . . , tn). The claim follows from the defi- nition of|P|and the list of primary mappings, since the ti are constant terms (or else P would not be closed).
(b2) P = ¬Q. If |P| = 1, then |Q| = 0 and Q ∈ E by the induction assumption applied toQ; sinceE is complete, we have¬Q∈ E, i.e.,P ∈ E. On the other hand, if|P|= 0, then|Q|= 1 andQ∈ E, so that ¬Q∈ E sinceE is consistent.
(b3)P = (Q1⇒Q2). We first show that if |P|= 0 then P ∈ E. In fact, in this case|Q1|= 1 and|Q2|= 0; by the induction assumption,Q1∈ E, Q2∈ E; sinceE is complete,¬Q2∈ E; using the tautologyQ1⇒(¬Q2⇒ ¬(Q1⇒Q2)) and using MP twice yieldsE (Q1⇒Q2). SinceE is complete and consistent, all closed formulas that are deducible from E belong to E; hence, ¬(Q1 ⇒ Q2) =¬P∈ E, so thatP∈ E.
We now show that if P ∈ E, then |P| = 0. In fact, since E is complete, we then have ¬P = ¬(Q1 ⇒ Q2) ∈ E. The tautologies ¬(Q1 ⇒ Q2) ⇒ Q1
and ¬(Q1 ⇒ Q2) ⇒ ¬Q2 and MP giveE Q1 andE¬Q2, so that since E is complete and consistent, Q1∈ E and ¬Q2 ∈ E. By the induction assumption,
|Q1|= 1 and |Q2|= 0, so that|P|=|Q1⇒Q2|= 0.
(b4)P =Q1∨Q2 orQ1∧Q2. Using the tautologies that express∧and∨ in terms of⇒and¬, we can reduce to the previous cases; we omit the details.
(b5)P =∀xQ. Ifxdoes not occur freely inQ, then|P|= 1 is equivalent to
|Q|= 1, i.e., by the induction assumption, to Q∈ E. ButQ∈ E is equivalent to ∀x Q∈ E, in one direction using Gen and in the other direction using the axiom of specialization witht=xand then MP.
We now assume that x occurs freely in Q. We first suppose that |P| = 1 butP ∈ E, and obtain a contradiction. IfP ∈ E, then¬P ∈ E, i.e.,¬∀x Q(x)∈ E. Since the alphabet of L is sufficient for E, it follows that E contains the formula ¬∀x Q(x) ⇒ ¬Q(cQ). Applying MP, we obtain E ¬Q(cQ); since E is consistent, we have Q(cQ) ∈ E. By the induction assumption, |Q(cQ)| = 0 (Q(cQ) is closed!). This means that |Q(x)|(ξ) = 0 for ξ ∈ M if xξ = cQ, contradicting the assumption that|P|= 1.
We now suppose that|P|= 0 butP ∈ E, and obtain a contradiction. Since
|P|= 0, for someξ∈M we have|Q(x)|(ξ) = 0. Let tbe the constant term for whichxξ =t. Clearlytis free forxinQ, so that 0 =|Q(x)|(ξ) =|Q(t)|. Hence Q(t) ∈ E by the induction assumption, and ¬Q(t) ∈ E since E is complete.
On the other hand, ifP ∈ E, i.e.,∀x Qx∈ E, then the axiom of specialization
∀x Q(x)⇒Q(t) gives us E Q(t). But since¬Q(t)∈ E, this contradicts the consistency ofE.
(b6) P = ∃x Q. This reduces to the previous case using the axiom that expresses∃in terms of∀ and negation; we omit the details.
6.10.Proof Of Lemma6.6. In order to embedEin a complete and consistent set E, we shall have to use Zorn’s lemma and the deduction lemma forL(see Section 4.5 of Chapter II). Zorn’s lemma will be applied to the set CE = the set of sets of formulasE inLthat containE and are consistent. The setCE is ordered by inclusion.
Verification of the hypothesis of Zorn’s lemma. Let{Eα}α∈I be a lin- early ordered subset of CE, i.e., for any α and β we have either Eα Eβ or Eβ Eα. Then the union∪Eα a belongs toCE. In fact, otherwise∪Eα would be inconsistent, and there would exist a deduction of a contradiction from a finite number of formulas. Suppose these formulas are contained inEα1, . . . ,Eαn. But one of these sets contains the remainingn−1; this set would be inconsistent, contrary to the definition of CE.
Proof of lemma 6.6 from Zorn’s lemma. The set CE has a maximal element, i.e., a consistent set E ⊃ E such that if Q ∈ E then E ∪ {Q} is inconsistent. We claim that E is complete. In fact, suppose that there were a closed formula P such that P ∈ E and ¬P ∈ E. Since E is maximal, it follows that E ∪ {P} R and E ∪ {¬P} R for any formula R. By the deduction lemma, E P ⇒ R and E ¬P ⇒ R. Using the tautology (P ⇒ R) ⇒ ((¬P ⇒ R) ⇒ R)) and MP, we have E R, contradicting the
consistency ofE.
6.11. Proof of Lemma6.7. In constructing a language with a sufficient alpha- bet for a consistent set of formulas E that containsE and AxL, we proceed in the most natural way.
(a) We add to the alphabet ofLa set of new constants whose cardinality is that of the alphabet ofL+ℵ0. We obtain a languageL.
(b) We consider the set of formulas E ∪ Ax L in the language L, where Ax L consists of all the logical axioms ofL. We claim that this set of formulas is consistent. In fact, if there were a deduction of a contradiction fromE ∪AxL in L, then the following procedure would transform it into a deduction of a contradiction fromE inL: take the finite set consisting of all the new constants that occur in the formulas in the deduction and replace these constants by old variables (in L) that do not occur in the formulas in the deduction. It is easily verified that the deduction of a contradiction remains a deduction of a contradiction, and now lies entirely inL.
(c) We consider the setS of formulas P(x) containing one free variable x and such that ¬∀xP(x) ∈ E ∪ Ax L. For each P(x) in S we choose a new constant cP subject to the following restriction: each cP can be assigned a natural number, itsrank, in such a way that if a constant of ranknoccurs in P(x) thencP has rank> n. This can be done since card(S)card(alphabet of L) = card(alphabet of L) +ℵ0. For eachP(x) inS define the formula
Rp:¬∀x P(x) ⇒ ¬P(cP) and finally let
E =E ∪Ax L∪ {RP|P(x)∈S}.
Call anyRP anR-formula. Note that noR-formula has the form¬∀x P(x), so that L is sufficient forE. It remains only to verify that E is consistent. If a contradiction were deducible from E then it would be deducible using finitely manyR-formulas. At least oneRP among these must be such thatcP does not occur in any of the others: namely, pick cP of maximal rank. Hence it suffices to verify that if E ∪AxL ∪ Ris consistent, where Ris a set of formulas not containingcP, then the addition ofRP does not lead to a contradiction.
SupposeE ∪Ax L∪ R ∪ {Rp} were inconsistent. Then, in particular, we would have a deduction of¬RP and, by the deduction lemma,E ∪AxL∪ R RP ⇒ ¬RP. The tautology (RP ⇒ ¬RP) ⇒ ¬RP and MP would yield a deduction of¬RP; that is,
E ∪Ax L∪ R (¬∀x P(x)⇒ ¬P(cP)).
Then the tautology ¬(P ⇒ ¬Q) ⇒ Q and MP would yield a deduction of P(cp). Transform this deduction by replacing the constantcP with a variable y that does not occur in the formulas in the deduction. Since cP does not occur in R it is easily verified that the transformation yields a deduction of P(y) from E ∪ AxL ∪ R. Using Gen, E ∪AxL ∪ R ∀y P(y). But since
¬∀x P(x)∈ E ∪AxL, we haveE ∪AxL ¬∀y P(y). Hence E ∪AxL∪ Ris
inconsistent, contrary to hypothesis.
6.12. Proof Of Lemma 6.8. Let L be a language in the classL1, and letE be a set of formulas in L. We embed E in a complete and consistent set E, and then apply Lemma 6.7 to (L,E). We let L∗ and E∗ denote the resulting language and set of formulas. We further define inductively
(L(0),E(0)) = (L,E), (L(i+1),E(i+1)) = (L(i)∗,E(i)∗),
and finally
L(∞)= ∞ i=0
L(i), E(∞)= ∞ i=0
E(i).
The setE(∞) is consistent, since any deduction of a contradiction would be obtained “at some finite level,” and all theE(i) are consistent. It is complete, since every closed formula inL(∞)is written in the alphabet ofL(i)for somei, andE(i+1)contains the completion ofE(i)inL(i). Finally, the alphabet ofL(∞) is sufficient forE(∞)by the same argument.
This completes the proof of the lemmas.
6.13.Deduction of theorem 6.2 from the lemmas. LetT be a G¨odelian set of formulas inL. Applying Lemma 6.8 toT, we embed (L, T) in (L(∞), T(∞)), where the pair (L∞), T(∞)) satisfies Lemma 6.5. Letφ(∞)be an interpretation of L(∞) such as must exist by Lemma 6.5. The cardinality of M(∞) does not exceed card(alphabet ofL)+ℵ0. The restrictionφofφ(∞)toLsatisfies the con- dition T ⊂TφL. We prove thatT =TφL. In fact, letP ∈TφL. If P is closed, then P ∈ T, since either P or ¬P lies in T by completeness, and ¬P ∈ T because P is φ-true. IfP is not closed, and x1, . . . , xn are the variables that occur freely inP, then∀xnP is closed and belongs toT. By the axiom of spe- cialization,P is deducible from T∪ {∀x1· · · ∀xnP}, so thatP ∈T, sinceT is closed under deduction. This proves the first assertion of the theorem.
The second assertion follows from the analogous argument applied to E instead of T. We find a model φ for E; then E ⊂ TφL and TφL is
G¨odelian.
6.14. In conclusion, we note that if the alphabet of L contains a symbol = for which the axioms of equality are included in E (or T), then there exists a normal interpretation that satisfies Theorem 6.2 and takes = into equality.
To prove this, we take the above model M and divide out by the equivalence relationφ(=), as in Section 4.6.