truth-value semantics for intuitionistic propositional logic is sound

truth-value semantics for intuitionistic propositional logic is sound

Proposition 1

.

The

truth-value semantics for intuitionistic propositional logic is sound

.

Proof.

We show that, for each

positive integer

n

, every

theorem

of intuitionistic

propositional logic

is a

tautology

for

V

n

. This amounts to showing that, for every

interpretation

v

on

V

n

,

•

each axiom is true, and

•

modus ponens

preserves truth.

Let us take care of the second one first. Suppose

v

(

A

)

=

v

(

A

→

B

)

=

n

. If

v

⁢

(

A

)

≤

v

⁢

(

B

)

, then

v

⁢

(

B

)

=

n

. Otherwise,

v

⁢

(

B

)

<

v

⁢

(

A

)

. But this means that

n

=

v

(

A

→

B

)

=

v

(

B

)

,

forcing

v

⁢

(

B

)

=

n

. Therefore,

v

⁢

(

B

)

=

n

.

Now, we verify that each of the

axiom schemas

below are true:

1.

(

A

∧

B

)

→

A

and

(

A

∧

B

)

→

B

.

Since

v

⁢

(

A

∧

B

)

=

min

⁡

{

v

⁢

(

A

)

,

v

⁢

(

B

)

}

≤

v

⁢

(

A

)

, we get

v

(

(

A

∧

B

)

→

A

)

=

n

. The other one is proved similarly.

2.

A

→

(

A

∨

B

)

and

B

→

(

A

∨

B

)

.

Since

v

⁢

(

A

)

≤

max

⁡

{

v

⁢

(

A

)

,

v

⁢

(

B

)

}

=

v

⁢

(

A

∨

B

)

, we get

v

(

A

→

(

A

∨

B

)

)

=

n

. The other one is proved similarly.

3.

A

→

(

B

→

A

)

.

If

v

⁢

(

B

)

≤

v

⁢

(

A

)

,

v

(

B

→

A

)

=

n

, so that

v

(

A

→

(

B

→

A

)

)

=

n

as well. If

v

⁢

(

A

)

<

v

⁢

(

B

)

, then

v

(

B

→

A

)

=

v

(

A

)

, so that

v

(

A

→

(

B

→

A

)

)

=

n

.

4.

¬

A

→

(

A

→

B

)

.

If

v

⁢

(

A

)

≤

v

⁢

(

B

)

,

v

(

A

→

B

)

=

n

, so that

v

(

¬

A

→

(

A

→

B

)

)

=

n

as well. If

v

⁢

(

B

)

<

v

⁢

(

A

)

, then

v

(

A

→

B

)

=

v

(

B

)

. Also,

v

⁢

(

B

)

<

v

⁢

(

A

)

implies that

v

⁢

(

A

)

>

0

, so that

v

⁢

(

¬

⁢

A

)

=

0

, and

v

(

¬

A

→

(

A

→

B

)

)

=

n

as a result.

5.

A

→

(

B

→

(

A

∧

B

)

)

.

If

v

⁢

(

B

)

=

v

⁢

(

A

∧

B

)

, then

v

⁢

(

B

)

≤

v

⁢

(

A

)

and

v

(

B

→

(

A

∧

B

)

)

=

n

, so that

v

(

A

→

(

B

→

(

A

∧

B

)

)

=

n

also. If on the other hand

v

⁢

(

A

∧

B

)

<

v

⁢

(

B

)

, then

v

(

A

)

=

v

(

A

∧

B

)

and

v

(

B

→

(

A

∧

B

)

)

=

v

(

A

∧

B

)

,

so that

v

(

A

→

(

B

→

(

A

∧

B

)

)

)

=

v

(

A

→

(

A

∧

B

)

)

=

n

.

6.

(

A

→

C

)

→

(

(

B

→

C

)

→

(

(

A

∨

B

)

→

C

)

)

.

If

v

⁢

(

B

)

=

v

⁢

(

A

∨

B

)

, then

v

⁢

(

A

)

≤

v

⁢

(

B

)

, and

v

(

(

B

→

C

)

→

(

(

A

∨

B

)

→

C

)

)

=

v

(

(

B

→

C

)

→

(

B

→

C

)

)

=

n

,

so that

v

(

(

A

→

C

)

→

(

(

B

→

C

)

→

(

(

A

∨

B

)

→

C

)

)

)

=

n

as well. Otherwise,

v

⁢

(

B

)

<

v

⁢

(

A

)

=

v

⁢

(

A

∨

B

)

. This means that

v

(

(

B

→

C

)

→

(

(

A

∨

B

)

→

C

)

)

)

=

v

(

(

B

→

C

)

→

(

A

→

C

)

)

,

and therefore

v

(

(

A

→

C

)

→

(

(

B

→

C

)

→

(

(

A

∨

B

)

→

C

)

)

)

=

v

(

(

A

→

C

)

→

(

(

B

→

C

)

→

(

A

→

C

)

)

)

=

n

by 3 above.

7.

(

A

→

B

)

→

(

(

A

→

(

B

→

C

)

)

→

(

A

→

C

)

)

.

It is clear that

v

(

C

)

≤

v

(

B

→

C

)

. If

v

(

C

)

=

v

(

B

→

C

)

, then

v

(

(

A

→

(

B

→

C

)

)

→

(

A

→

C

)

)

=

v

(

(

A

→

C

)

→

(

A

→

C

)

)

=

n

,

so that

v

(

(

A

→

B

)

→

(

(

A

→

(

B

→

C

)

)

→

(

A

→

C

)

)

)

=

n

too. If

v

(

C

)

<

v

(

B

→

C

)

, then

v

(

B

→

C

)

=

n

, which implies

v

⁢

(

B

)

≤

v

⁢

(

C

)

. This in turn implies that

v

(

A

→

B

)

≤

v

(

A

→

C

)

, so that

v

(

(

A

→

C

)

→

(

(

A

→

(

B

→

C

)

)

→

(

A

→

C

)

)

)

≤

v

(

(

A

→

B

)

→

(

(

A

→

(

B

→

C

)

)

→

(

A

→

C

)

)

)

.

But by 3 above,

v

(

(

A

→

C

)

→

(

(

A

→

(

B

→

C

)

)

→

(

A

→

C

)

)

)

=

n

,

hence

v

(

(

A

→

B

)

→

(

(

A

→

(

B

→

C

)

)

→

(

A

→

C

)

)

)

=

n

as a result.

8.

(

A

→

B

)

→

(

(

A

→

¬

B

)

→

¬

A

)

.

Pick any

C

such that

v

⁢

(

C

)

=

0

, such as

D

∧

¬

⁢

D

. Then

v

(

¬

B

)

=

v

(

B

→

C

)

, so that

v

(

(

A

→

B

)

→

(

(

A

→

¬

B

)

→

¬

A

)

)

=

v

(

(

A

→

B

)

→

(

(

A

→

(

B

→

C

)

)

→

(

A

→

C

)

)

)

=

n

by 7.

∎

Note that the proofs of the axioms employ some

elementary

facts, for any wff’s

A

,

B

,

C

:

•

If

v

⁢

(

B

)

=

n

or

v

⁢

(

A

)

=

0

, then

v

(

A

→

B

)

=

n

.

•

if

v

⁢

(

B

)

=

0

, then

v

(

A

→

B

)

=

v

(

¬

A

)

.

•

if

v

⁢

(

A

)

=

n

, then

v

(

A

→

B

)

=

v

(

B

)

.

•

v

(

B

)

≤

v

(

A

→

B

)

.

•

if

v

⁢

(

B

)

≤

v

⁢

(

C

)

, then

–

v

⁢

(

A

∨

B

)

≤

v

⁢

(

A

∨

C

)

,

–

v

⁢

(

A

∧

B

)

≤

v

⁢

(

A

∧

C

)

,

–

v

(

A

→

B

)

≤

v

(

A

→

C

)

,

–

v

(

C

→

A

)

≤

v

(

B

→

A

)

.

From the facts above, one readily deduces:

•

if

v

⁢

(

B

)

≤

v

⁢

(

C

)

, then

v

⁢

(

¬

⁢

C

)

≤

v

⁢

(

¬

⁢

B

)

,

•

if

v

⁢

(

B

)

=

v

⁢

(

C

)

, then

–

v

⁢

(

A

∨

B

)

=

v

⁢

(

A

∨

C

)

,

–

v

⁢

(

A

∧

B

)

=

v

⁢

(

A

∧

C

)

,

–

v

(

A

→

B

)

=

v

(

A

→

C

)

,

–

v

(

C

→

A

)

=

v

(

B

→

A

)

,

–

v

⁢

(

¬

⁢

B

)

=

v

⁢

(

¬

⁢

C

)

.

Title

truth-value semantics for intuitionistic propositional logic is sound

Canonical name

TruthvalueSemanticsForIntuitionisticPropositionalLogicIsSound

Date of creation

2013-03-22 19:31:38

Last modified on

2013-03-22 19:31:38

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

21

Author

CWoo (3771)

Entry type

Definition

Classification

msc 03B20

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