Download PDF

properties of consistency∗
CWoo†
2013-03-22 4:01:31
Fix a (classical) propositional logic L. Recall that a set ∆ of wff’s is said
to be L-consistent, or consistent for short, if ∆ 6`⊥. In other words, ⊥ can not
be derived from axioms of L and elements of ∆ via finite applications of modus
ponens. There are other equivalent formulations of consistency:
1. ∆ is consistent
2. Ded(∆) := {A | ∆ ` A} is not the set of all wff’s
3. there is a formula A such that ∆ 6` A.
4. there are no formulas A such that ∆ ` A and ∆ ` ¬A.
Proof. We shall prove 1. ⇒ 2. ⇒ 3. ⇒ 4. ⇒ 1.
1. ⇒ 2. Since ⊥∈
/ {A | ∆ ` A}.
2. ⇒ 3. Any formula not in {A | ∆ ` A} will do.
3. ⇒ 4. If ∆ ` A and ∆ ` ¬A, then A, A →⊥, ⊥, ⊥→ B, B is a deduction of B
from A and ¬A, but this means that ∆ ` B for any wff B.
4. ⇒ 1. Since ∆ ` ¬ ⊥, ∆ 6`⊥ as a result.
Below are some properties of consistency:
1. ∆ ∪ {A} is consistent iff ∆ 6` ¬A.
2. ∆ ∪ {¬A} is not consistent iff ∆ ` A.
3. Any subset of a consistent set is consistent.
4. If ∆ is consistent, so is Ded(∆).
∗ hPropertiesOfConsistencyi created: h2013-03-2i by: hCWooi version: h42574i Privacy
setting: h1i hFeaturei h03B45i h03B10i h03B05i h03B99i
† This text is available under the Creative Commons Attribution/Share-Alike License 3.0.
You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
1
5. If ∆ is consistent, then at least one of ∆ ∪ {A} or ∆ ∪ {¬A} is consistent
for any wff A.
6. If there is a truth-valuation v such that v(A) = 1 for all A ∈ ∆, then ∆ is
consistent.
7. If 6` A, and ∆ contains the schema based on A, then ∆ is not consistent.
Remark. The converse of 6 is also true; it is essentially the compactness
theorem for propositional logic (see here).
Proof. The first two are contrapositive of one another via the theorem A ↔
¬¬A, so we will just prove one of them.
2. ∆, ¬A `⊥ iff ∆ ` ¬¬A by the deduction theorem iff ∆ ` A by the
substitution theorem.
3. If Γ is not consistent, Γ `⊥. If Γ ⊆ ∆, then ∆ `⊥ as well, so ∆ is not
consistent.
4. Since ∆ is consistent, ⊥∈
/ Ded(∆). Now, if Ded(∆) `⊥, but by the remark
below, ⊥∈ Ded(∆), a contradiction.
5. Suppose ∆ is consistent and A any wff. If neither ∆ ∪ {A} and ∆ ∪ {¬A}
are consistent, then ∆, A `⊥ and ∆, ¬A `⊥, or ∆ ` ¬A and ∆ ` ¬¬A, or
∆ ` ¬A and ∆ ` A by the substitution theorem on A ↔ ¬¬A, but this
means ∆ is not consistent, a contradiction.
6. If v(A) = 1 for all A ∈ ∆, v(B) = 1 for all B such that ∆ ` B. Since
v(⊥) = 0, ∆ is consistent.
7. Suppose v(A) for some valuation v. Let p1 , . . . , pm be the propositional
variables in A such that v(pi ) = 0 and q1 , . . . , qn be the variables in A
such that v(qj ) = 1. Let A0 be the instance of the schema A where each pi
is replaced by ⊥ and each qj replaced by > (which is ¬ ⊥). Then A0 ∈ ∆.
Furthermore, v(A0 ) = v(A) = 0. Now, for any valuation u, since u(⊥) = 0
and u(>) = 1, we get u(A0 ) = v(A0 ) = 0. In other words, u(¬A0 ) = 1
for all valuations u, so ¬A0 is valid, and hence a theorem of L by the
completeness theorem. But this means that A0 ↔⊥, which implies that
∆ `⊥.
Remark. The set Ded(∆) is called the deductive closure of ∆. It is so called
because it is deductively closed: A ∈ Ded(∆) iff Ded(∆) ` A.
Proof. If A ∈ Ded(∆), then ∆ ` A, so certainly Ded(∆) ` A, as Ded(∆) is a
superset of ∆.
Before proving the converse, note first that if ∆ ` B and ∆ ` B → A, ∆ ` A
by modus ponens. This implies that Ded(∆) is closed under modus ponens: if
B and B → A are both in Ded(∆), so is A.
2
Now, suppose Ded(∆) ` A. We induct on the length of the deduction
sequence of A. If n = 1, then A ∈ Ded(∆) and we are done. Now, suppose the
length of is n + 1. If A is either a theorem or in Ded(∆), we are done. Now,
suppose A is the result of applying modus ponens to two earlier members, say
Ai and Aj . Since A1 , . . . , Ai is a deduction of Ai from Ded(∆), and it has length
i ≤ n, by the induction step, Ai ∈ Ded(∆). Similarly, Aj ∈ Ded(∆). But this
means that A ∈ Ded(∆) by the last paragraph.
3