properties of consistency

properties of consistency

Fix a (classical)

propositional logic

L

. Recall that a set

Δ

of wff’s is said to be

L

-

consistent

, or

consistent

for short, if

Δ

⊬

⟂

. In other words,

⟂

can not be derived from axioms of

L

and elements of

Δ

via finite applications of

modus ponens

. There are other

equivalent

formulations of

consistency

:

1.

Δ

is consistent

2.

Ded

(

Δ

)

:=

{

A

∣

Δ

⊢

A

}

is not the set of all wff’s

3.

there is a

formula

A

such that

Δ

⊬

A

.

4.

there are no formulas

A

such that

Δ

⊢

A

and

Δ

⊢

¬

⁢

A

.

Proof.

We shall prove

1

.

⇒

2

.

⇒

3

.

⇒

4

.

⇒

1

.

1

.

⇒

2

.

Since

⟂

∉

{

A

∣

Δ

⊢

A

}

.

2

.

⇒

3

.

Any formula not in

{

A

∣

Δ

⊢

A

}

will do.

3

.

⇒

4

.

If

Δ

⊢

A

and

Δ

⊢

¬

⁢

A

, then

A

,

A

→

⟂

,

⟂

,

⟂

→

B

,

B

is a

deduction

of

B

from

A

and

¬

⁢

A

, but this means that

Δ

⊢

B

for any wff

B

.

4

.

⇒

1

.

Since

Δ

⊢

¬

⟂

,

Δ

⊬

⟂

as a result.

∎

Below are some

properties of consistency

:

1.

Δ

∪

{

A

}

is consistent iff

Δ

⊬

¬

⁢

A

.

2.

Δ

∪

{

¬

⁢

A

}

is not consistent iff

Δ

⊢

A

.

3.

Any subset of a consistent set is consistent.

4.

If

Δ

is consistent, so is Ded

(

Δ

)

.

5.

If

Δ

is consistent, then at least one of

Δ

∪

{

A

}

or

Δ

∪

{

¬

⁢

A

}

is consistent for any wff

A

.

6.

If there is a truth-valuation

v

such that

v

⁢

(

A

)

=

1

for all

A

∈

Δ

, then

Δ

is consistent.

7.

If

⊬

A

, and

Δ

contains the schema based on

A

, then

Δ

is not consistent.

Remark

. The

converse

of 6 is also true; it is essentially the

compactness theorem

for propositional logic (see here (

http://planetmath.org/CompactnessTheoremForClassicalPropositionalLogic

)).

Proof.

The first two are

contrapositive

of one another via the

theorem

A

↔

¬

⁢

¬

⁢

A

, so we will just prove one of them.

2.

Δ

,

¬

⁢

A

⊢

⟂

iff

Δ

⊢

¬

⁢

¬

⁢

A

by the

deduction theorem

iff

Δ

⊢

A

by the

substitution theorem

.

3.

If

Γ

is not consistent,

Γ

⊢

⟂

. If

Γ

⊆

Δ

, then

Δ

⊢

⟂

as well, so

Δ

is not consistent.

4.

Since

Δ

is consistent,

⟂

∉

Ded

(

Δ

)

. Now, if Ded

(

Δ

)

⊢

⟂

, but by the remark below,

⟂

∈

Ded

(

Δ

)

, a

contradiction

.

5.

Suppose

Δ

is consistent and

A

any wff. If neither

Δ

∪

{

A

}

and

Δ

∪

{

¬

⁢

A

}

are consistent, then

Δ

,

A

⊢

⟂

and

Δ

,

¬

⁢

A

⊢

⟂

, or

Δ

⊢

¬

⁢

A

and

Δ

⊢

¬

⁢

¬

⁢

A

, or

Δ

⊢

¬

⁢

A

and

Δ

⊢

A

by the substitution theorem on

A

↔

¬

⁢

¬

⁢

A

, but this means

Δ

is not consistent, a contradiction.

6.

If

v

⁢

(

A

)

=

1

for all

A

∈

Δ

,

v

⁢

(

B

)

=

1

for all

B

such that

Δ

⊢

B

. Since

v

⁢

(

⟂

)

=

0

,

Δ

is consistent.

7.

Suppose

v

⁢

(

A

)

for some

valuation

v

. Let

p

1

,

…

,

p

m

be the

propositional variables

in

A

such that

v

⁢

(

p

i

)

=

0

and

q

1

,

…

,

q

n

be the

variables

in

A

such that

v

⁢

(

q

j

)

=

1

. Let

A

′

be the

instance

of the schema

A

where each

p

i

is replaced by

⟂

and each

q

j

replaced by

⊤

(which is

¬

⟂

). Then

A

′

∈

Δ

. Furthermore,

v

⁢

(

A

′

)

=

v

⁢

(

A

)

=

0

. Now, for any valuation

u

, since

u

⁢

(

⟂

)

=

0

and

u

⁢

(

⊤

)

=

1

, we get

u

⁢

(

A

′

)

=

v

⁢

(

A

′

)

=

0

. In other words,

u

⁢

(

¬

⁢

A

′

)

=

1

for all valuations

u

, so

¬

⁢

A

′

is valid, and hence a theorem of

L

by the

completeness theorem

. But this means that

A

′

↔

⟂

, which implies that

Δ

⊢

⟂

.

∎

Remark

. The set Ded

(

Δ

)

is called the

deductive closure

of

Δ

. It is so called because it is

deductively closed

:

A

∈

Ded

⁢

(

Δ

)

iff Ded

(

Δ

)

⊢

A

.

Proof.

If

A

∈

Ded

⁢

(

Δ

)

, then

Δ

⊢

A

, so certainly Ded

(

Δ

)

⊢

A

, as Ded

(

Δ

)

is a

superset

of

Δ

.

Before proving the converse, note first that if

Δ

⊢

B

and

Δ

⊢

B

→

A

,

Δ

⊢

A

by modus ponens. This implies that Ded

(

Δ

)

is

closed under

modus ponens: if

B

and

B

→

A

are both in Ded

(

Δ

)

, so is

A

.

Now, suppose Ded

(

Δ

)

⊢

A

. We induct on the length of the deduction sequence of

A

. If

n

=

1

, then

A

∈

Ded

⁢

(

Δ

)

and we are done. Now, suppose the length of is

n

+

1

. If

A

is either a theorem or in Ded

(

Δ

)

, we are done. Now, suppose

A

is the result of applying modus ponens to two earlier members, say

A

i

and

A

j

. Since

A

1

,

…

,

A

i

is a deduction of

A

i

from Ded

(

Δ

)

, and it has length

i

≤

n

, by the

induction

step,

A

i

∈

Ded

⁢

(

Δ

)

. Similarly,

A

j

∈

Ded

⁢

(

Δ

)

. But this means that

A

∈

Ded

⁢

(

Δ

)

by the last paragraph.
∎

Title

properties of consistency

Canonical name

PropertiesOfConsistency

Date of creation

2013-03-22 19:35:07

Last modified on

2013-03-22 19:35:07

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

18

Author

CWoo (3771)

Entry type

Feature

Classification

msc 03B45

Classification

msc 03B10

Classification

msc 03B05

Classification

msc 03B99

Related topic

FirstOrderTheories

Defines

deductive closure

\@unrecurse

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