completeness theorem for propositional logic

completeness theorem for propositional logic

The

completeness theorem

of

propositional logic

is the statement that a wff is

tautology

iff it is a

theorem

. In symbol, we have

⊧

A

iff

⊢

A

for any wff

A

. The “if” part of the statement is the soundness theorem, and is proved here (

http://planetmath.org/TruthValueSemanticsForPropositionalLogicIsSound

). We will prove the “only if” part, which is also known as the completeness portion of the theorem. We will give a

constructive proof

of this. Before proving this, we state and prove some preliminary facts:

1.

A

,

B

⊢

A

→

B

2.

A

,

¬

⁢

B

⊢

¬

⁡

(

A

→

B

)

3.

¬

⁢

A

,

B

⊢

A

→

B

4.

¬

⁢

A

,

¬

⁢

B

⊢

A

→

B

5.

Let

v

be a

valuation

. For any wff

A

, let

v

⁢

[

A

]

be defined as follows:

v

⁢

[

A

]

⁢

is

⁢

{

A

if

⁢

v

⁢

(

A

)

=

1

,

¬

⁢

A

if

⁢

v

⁢

(

A

)

=

0

.

Suppose

p

1

,

…

,

p

n

are the

propositional variables

in

A

. Then

v

⁢

[

p

1

]

,

…

,

v

⁢

[

p

n

]

⊢

v

⁢

[

A

]

.

6.

if

Δ

,

A

⊢

B

and

Δ

,

¬

⁢

A

⊢

B

, then

⊢

B

.

Proof.

Facts 1 and 3 come from the

axiom schema

B

→

(

A

→

B

)

. From

⊢

B

→

(

A

→

B

)

, we have

C

⊢

B

→

(

A

→

B

)

, so

C

,

B

⊢

A

→

B

. If

C

is

A

, we have fact 1, and if

C

is

¬

⁢

A

, we have fact 3.

Fact 2: this is proved here (

http://planetmath.org/SubstitutionTheoremForPropositionalLogic

).

Fact 4: By ex falso quodlibet,

⊢

⟂

→

B

, so

A

⊢

⟂

→

B

, and therefore

⊢

A

→

(

⟂

→

B

)

by the

deduction theorem

. Now,

(

A

→

(

⟂

→

B

)

)

→

(

(

A

→

⟂

)

→

(

A

→

B

)

)

is an axiom

instance

, so

⊢

(

A

→

⟂

)

→

(

A

→

B

)

, or

⊢

¬

A

→

(

A

→

B

)

, or

¬

⁢

A

⊢

A

→

B

, and

¬

⁢

A

,

¬

⁢

B

⊢

A

→

B

all the more so.

Fact 5: by

induction

on the number

n

of

occurrences

of

→

in

A

. If

n

=

0

, then

A

is either

⟂

or a propositional variable

p

. In the first case,

v

⁢

[

A

]

is

¬

⟂

, and from

⟂

⁣

⊢

⁣

⟂

, we get

⊢

⁣

⟂

⁣

→

⁣

⟂

, or

⊢

¬

⟂

. In the second case,

v

⁢

[

p

]

⊢

v

⁢

[

p

]

. Now, suppose there are

n

+

1

occurrences of

→

in

A

. Let

p

1

,

…

,

p

m

be the propositional variables in

A

. By

unique readability

,

A

is

B

→

C

for some unique wff’s

B

and

C

. Since each

B

and

C

has no more than

n

occurrences of

→

, by induction, we have

v

⁢

[

p

i

⁢

(

1

)

]

,

…

,

v

⁢

[

p

i

⁢

(

s

)

]

⊢

v

⁢

[

B

]

and

v

⁢

[

p

j

⁢

(

1

)

]

,

…

,

v

⁢

[

p

j

⁢

(

t

)

]

⊢

v

⁢

[

C

]

,

where the propositional variables in

B

are

p

i

⁢

(

1

)

,

…

,

p

i

⁢

(

s

)

, and in

C

are

p

j

⁢

(

1

)

,

…

,

p

j

⁢

(

t

)

. So

v

⁢

[

p

1

]

,

…

,

v

⁢

[

p

m

]

⊢

v

⁢

[

B

]

and

v

⁢

[

p

1

]

,

…

,

v

⁢

[

p

m

]

⊢

v

⁢

[

C

]

.

Next, we want to show that

v

[

B

]

,

v

[

C

]

⊢

v

[

B

→

C

]

. We break this into four cases:

•

if

v

⁢

[

B

]

is

B

and

v

⁢

[

C

]

is

C

: then

v

[

B

→

C

]

is

B

→

C

, and use Fact 1

•

if

v

⁢

[

B

]

is

B

and

v

⁢

[

C

]

is

¬

⁢

C

: then

v

[

B

→

C

]

is

¬

⁡

(

B

→

C

)

, and use Fact 2

•

if

v

⁢

[

B

]

is

¬

⁢

B

and

v

⁢

[

C

]

is

C

: then

v

[

B

→

C

]

is

B

→

C

, and use Fact 3

•

if

v

⁢

[

B

]

is

¬

⁢

B

and

v

⁢

[

C

]

is

¬

⁢

C

: then

v

[

B

→

C

]

is

B

→

C

, and use Fact 4.

In all cases, we have by applying

modus ponens

,

v

[

p

1

]

,

…

,

v

[

p

m

]

⊢

v

[

B

→

C

]

.

Fact 6 is proved here (

http://planetmath.org/SubstitutionTheoremForPropositionalLogic

).
∎

Theorem 1

.

Propositional logic is

complete

with respect to truth-value semantics.

Proof.

Suppose

A

is a tautology. Let

p

1

,

…

,

p

n

be the propositional variables in

A

. Then

v

⁢

[

p

1

]

,

…

,

v

⁢

[

p

n

]

⊢

v

⁢

[

A

]

for any valuation

v

. Since

v

⁢

[

A

]

is

A

. We have

v

⁢

[

p

1

]

,

…

,

v

⁢

[

p

n

]

⊢

A

.

If

n

=

0

, then we are done. So suppose

n

>

0

. Pick a valuation

v

such that

v

⁢

(

p

n

)

=

1

, and a valuation

v

′

such that

v

′

⁢

(

p

i

)

=

v

⁢

(

p

i

)

and

v

′

⁢

(

p

n

)

=

0

Then

v

⁢

[

p

1

]

,

…

,

v

⁢

[

p

n

-

1

]

,

p

n

⊢

A

and

v

⁢

[

p

1

]

,

…

,

v

⁢

[

p

n

-

1

]

,

¬

⁢

p

n

⊢

A

,

where the first

deducibility relation

comes from

v

and the second comes from

v

′

. By Fact 6 above,

v

⁢

[

p

1

]

,

…

,

v

⁢

[

p

n

-

1

]

⊢

A

.

So we have eliminated

v

⁢

[

p

n

]

from the left of

v

⁢

[

p

1

]

,

…

,

v

⁢

[

p

n

]

⊢

A

. Now, repeat this process until all of the

v

⁢

[

p

i

]

have been eliminated, and we have

⊢

A

.
∎

The completeness theorem can be used to show that certain complicated wff’s are theorems. For example, one of the

distributive laws

⊢

(

A

∧

B

)

∨

C

↔

(

A

∨

C

)

∧

(

B

∨

C

)

To see that this is indeed a theorem, by the completeness theorem, all we need to show is that it is true using the

truth table

:

(

A

∧

B

)

∨

C

↔

(

A

∨

C

)

∧

(

B

∨

C

)

T

T

T

T

T

T

T

T

T

T

T

T

T

T

T

T

T

F

T

T

T

F

T

T

T

F

T

F

F

T

T

T

T

T

T

T

F

T

T

T

F

F

F

F

T

T

T

F

F

F

F

F

F

F

T

T

T

T

F

T

T

T

T

T

T

F

F

T

F

F

T

F

F

F

F

T

T

F

F

F

F

T

T

T

F

T

T

T

F

T

T

F

F

F

F

F

T

F

F

F

F

F

F

F

S

⁢

i

⁢

m

⁢

i

⁢

l

⁢

a

⁢

r

⁢

l

⁢

y

,

o

⁢

n

⁢

e

⁢

c

⁢

a

⁢

n

⁢

s

⁢

h

⁢

o

⁢

w

⊢(A∨B)∧C ↔(A∧C)∨(B∧C)

.

Title

completeness theorem for propositional logic

Canonical name

CompletenessTheoremForPropositionalLogic1

Date of creation

2013-03-22 19:32:47

Last modified on

2013-03-22 19:32:47

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

21

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 03B05

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