2.6.4: Henkin Expansion
- Page ID
- 121713
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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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