2.6.4: Henkin Expansion
Part of the challenge in proving the completeness theorem is that the model we construct from a complete consistent set \(\Gamma\) must make all the quantified formulas in \(\Gamma\) true. In order to guarantee this, we use a trick due to Leon Henkin. In essence, the trick consists in expanding the language by infinitely many constant symbols and adding, for each formula with one free variable \(A(x)\) a formula of the form \(\lexists{x}{A(x)} \lif A(c)\) , where \(c\) is one of the new constant symbols. When we construct the structure satisfying \(\Gamma\) , this will guarantee that each true existential sentence has a witness among the new constants.
If \(\Gamma\) is consistent in \(\Lang L\) and \(\Lang L'\) is obtained from \(\Lang L\) by adding a countably infinite set of new constant symbols \(\Obj d_0\) , \(\Obj d_1\) , …, then \(\Gamma\) is consistent in \(\Lang L'\) .
A set \(\Gamma\) of formulas of a language \(\Lang {L}\) is saturated iff for each formula \(A(x) \in \Frm[L]\) with one free variable \(x\) there is a constant symbol \(c \in \Lang{L}\) such that \(\lexists{x}{A(x)} \lif A(c) \in \Gamma\) .
The following definition will be used in the proof of the next theorem.
Let \(\Lang L'\) be as in Proposition \(\PageIndex{1}\) . Fix an enumeration \(A_0(x_0)\) , \(A_1(x_1)\) , …of all formulas \(A_i(x_i)\) of \(\Lang L'\) in which one variable ( \(x_i\) ) occurs free. We define the sentences \(D_n\) by induction on \(n\) .
Let \(c_0\) be the first constant symbol among the \(\Obj d_i\) we added to \(\Lang{L}\) which does not occur in \(A_0(x_0)\) . Assuming that \(D_0\) , …, \(D_{n-1}\) have already been defined, let \(c_n\) be the first among the new constant symbols \(\Obj d_i\) that occurs neither in \(D_0\) , …, \(D_{n-1}\) nor in \(A_n(x_n)\) .
Now let \(D_{n}\) be the formula \(\lexists{x_{n}}{A_{n}(x_{n})} \lif A_{n}(c_{n})\) .
Every consistent set \(\Gamma\) can be extended to a saturated consistent set \(\Gamma'\) .
Proof. Given a consistent set of sentences \(\Gamma\) in a language \(\Lang{L}\) , expand the language by adding a countably infinite set of new constant symbols to form \(\Lang{L'}\) . By Proposition \(\PageIndex{1}\) , \(\Gamma\) is still consistent in the richer language. Further, let \(D_i\) be as in Definition \(\PageIndex{2}\) . Let \[\begin{aligned} \Gamma_0 & = \Gamma \\ \Gamma_{n+1} & = \Gamma_n \cup \{D_n \}\end{aligned}\] i.e., \(\Gamma_{n+1} = \Gamma \cup \{ D_0, \dots, D_n \}\) , and let \(\Gamma' = \bigcup_{n} \Gamma_n\) . \(\Gamma'\) is clearly saturated.
If \(\Gamma'\) were inconsistent, then for some \(n\) , \(\Gamma_n\) would be inconsistent (Exercise: explain why). So to show that \(\Gamma'\) is consistent it suffices to show, by induction on \(n\) , that each set \(\Gamma_n\) is consistent.
The induction basis is simply the claim that \(\Gamma_0 = \Gamma\) is consistent, which is the hypothesis of the theorem. For the induction step, suppose that \(\Gamma_{n}\) is consistent but \(\Gamma_{n+1} = \Gamma_n \cup \{D_n\}\) is inconsistent. Recall that \(D_n\) is \(\lexists{x_{n}}{A_{n}(x_n)} \lif A_{n}(c_{n})\) , where \(A_n(x_n)\) is a formula of \(\Lang{L'}\) with only the variable \(x_n\) free. By the way we’ve chosen the \(c_n\) (see Definition \(\PageIndex{2}\) ), \(c_n\) does not occur in \(A_n(x_n)\) nor in \(\Gamma_n\) .
If \(\Gamma_n \cup \{D_n\}\) is inconsistent, then \(\Gamma_n \Proves \lnot D_n\) , and hence both of the following hold: \[\Gamma_n \Proves \lexists{x_n}{A_n(x_n)} \qquad \Gamma_n \Proves \lnot A_n(c_n)\nonumber\] Since \(c_n\) does not occur in \(\Gamma_n\) or in \(A_n(x_n)\) , Theorems 8.11.1 and 9.10.1 applies. From \(\Gamma_n \Proves \lnot A_n(c_n)\) , we obtain \(\Gamma_n \Proves \lforall{x_n}{\lnot A_n(x_n)}\) . Thus we have that both \(\Gamma_n \Proves \lexists{x_n}{A_n(x_n)}\) and \(\Gamma_n \Proves \lforall{x_n}{\lnot A_n(x_n)}\) , so \(\Gamma_n\) itself is inconsistent. (Note that \(\lforall{x_n}{\lnot A_n(x_n)} \Proves \lnot\lexists{x_n}{A_n(x_n)}\) .) Contradiction: \(\Gamma_n\) was supposed to be consistent. Hence \(\Gamma_n \cup \{ D_n\}\) is consistent. ◻
We’ll now show that complete , consistent sets which are saturated have the property that it contains a universally quantified sentence iff it contains all its instances and it contains an existentially quantified sentence iff it contains at least one instance . We’ll use this to show that the structure we’ll generate from a complete, consistent, saturated set makes all its quantified sentences true.
Suppose \(\Gamma\) is complete, consistent, and saturated.
- \(\lexists{x}{A(x)} \in \Gamma\) iff \(A(t) \in \Gamma\) for at least one closed term \(t\) .
- \(\lforall{x}{A(x)} \in \Gamma\) iff \(A(t) \in \Gamma\) for all closed terms \(t\) .
Proof.
-
First suppose that
\(\lexists{x}{A(x)} \in \Gamma\)
. Because
\(\Gamma\)
is saturated,
\((\lexists{x}{A(x)} \lif A(c)) \in \Gamma\)
for some constant symbol
\(c\)
. By Propositions
8.10.3
and
9.9.3
, item (1), and Proposition
10.3.1
(1)
,
\(A(c) \in \Gamma\)
.
For the other direction, saturation is not necessary: Suppose \(A(t) \in \Gamma\) . Then \(\Gamma \Proves \lexists{x}{A(x)}\) by Propositions 8.11.1 and 9.10.1 , item (1). By Proposition 10.3.1 (1) , \(\lexists{x}{A(x)} \in \Gamma\) .
- Exercise.
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