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Lindenbaum’s lemma∗
CWoo†
2013-03-22 4:01:57
In this entry, we prove the following assertion known as Lindenbaum’s lemma:
every consistent set is a subset of a maximally consistent set. From this, we automatically conclude the existence of a maximally consistent set based on the
fact that the empty set is consistent (vacuously).
Proposition 1. (Lindenbaum’s lemma) Every consistent set can be extended
to a maximally consistent set.
We give two proofs of this, one uses Zorn’s lemma, and the other does not,
and is based on the countability of the set of wff’s in the logic. Both proofs
require the following:
Lemma 1. The union of a chain of consistent sets, ordered by ⊆, is consistent.
Proof. Let C := {Γi | i ∈ I} be a chain of consistent sets
S ordered by ⊆ and
indexed by some set I. We want to show that Γ := C is also consistent.
Suppose not. Then Γ `⊥. Let A1 , . . . , An be a deduction of ⊥ (which is An )
from Γ. Then for each j = 1, . . . , n − 1, there is some Γi(j) ∈ C such that Γi(j) `
Ai . Since C is a chain, take the largest of the Γi(j) ’s, say Γk , so that Γk ` Ai
for all i = 1, . . . , n − 1. This implies that Γk ` An , or Γk `⊥, contradicting the
assumption that Γk is consistent. As a result, Γ is consistent.
First Proof. Suppose ∆ is consistent. Let P be the partially ordered set of
all consistent
S supersets of ∆, ordered by inclusion ⊆. If
SC is a chain of elements
in P , then C is consistent by the lemma above, so C ∈ P as each element
of C is a superset of ∆. By Zorn’s lemma, P has a maximal element, call it Γ.
To see that Γ is maximally consistent, suppose Γ is not maximal. Then there is
a wff A such that Γ 6` A and Γ 6` ¬A, the first of which implies that A ∈
/ Γ, and
the second of which implies that Γ ∪ {A} is consistent, and therefore in P . The
two imply that Γ ∪ {A} is a consistent proper superset of Γ, contradicting the
maximality of Γ in P . Therefore, Γ is maximally consistent.
∗ hLindenbaumsLemmai created: h2013-03-2i by: hCWooi version: h42577i Privacy
setting: h1i hTheoremi h03B45i h03B10i h03B05i h03B99i
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You can reuse this document or portions thereof only if you do so under terms that are
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Second Proof. Let A1 , . . . , An , . . . be an enumeration of all wff’s of the logic
in question (this can be achieved if the set of propositional variables can be
enumerated). Let ∆ be a consistent set of wff’s. Define sets Γ1 , Γ2 , . . . , Γ of
wff’s inductively as follows:
Γ1
Γn+1
Γ
:=
∆
(
Γn ∪ {An }
:=
Γn ∪ {¬An }
:=
∞
[
if Γn ` An
otherwise
Γi .
i=1
First, notice that each Γi is consistent. This is done by induction on i. By
assumption, the case is true when i = 1. Now, suppose Γn is consistent. Then its
deductive closure Ded(Γn ) is also consistent. If Γn ` An , then clearly Γn ∪ {An }
is consistent since it is a subset of Ded(Γ). Otherwise, Γn 6` An , or Γn 6` ¬¬An
by the substitution theorem, and therefore Γn ∪ {¬An } by one of the properties
of consistency (see here). In either case, Γn+1 is consistent.
Next, Γ is maximally consistent. Γ is consistent because, by the lemma
above, it is the union of a chain of consistent sets. To see that Γ is maximal,
pick any wff A. Then A is some An in the enumerated list of all wff’s. Therefore,
either A ∈ Γn+1 or ¬A ∈ Γn+1 . Since Γn+1 ⊆ Γ, we have A ∈ Γ or ¬A ∈ Γ,
which implies that Γ is maximal (see here).
.
Given a logic L, let WL be the set of all maximally L-consistent sets. By
Lindebaum’s lemma, WL 6= ∅. We record two useful corollaries:
T
• For any consistent set ∆, Ded(∆) = {Γ ∈ WL | ∆ ⊆ Γ}.
T
• L = WL .
Proof. The second statement is a corollary of the first, for L = Ded(∅). To see
the first, let D := {Γ ∈ WL | ∆ ⊆ Γ}. Then D 6= ∅ by Lindebaum’s lemma.
Also, for any Γ ∈ D, Ded(∆) ⊆ Γ since Γ is deductively closed. On the other
hand if A ∈
/ Ded(∆), then ∆ 6` A, so ∆ ∪ {¬A} is consistent, and therefore is
contained in a maximally consistent
set Γ0 ∈ D by Lindenbaum’s lemma. Since
T
0
0
¬A ∈ Γ , A ∈
/ Γ , so that A ∈
/ D.
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