deduction theorem holds for intuitionistic propositional logic

deduction theorem holds for intuitionistic propositional logic

In this entry, we show that the

deduction theorem

below holds for intuitionistic

propositional logic

. We use the

axiom system

provided in this entry (

http://planetmath.org/AxiomSystemForIntuitionisticLogic

).

Theorem 1

.

If

Δ

,

A

⊢

i

B

, where

Δ

is a set of wff’s of the intuitionistic propositional logic, then

Δ

⊢

i

A

→

B

.

The proof is very

similar

to that of the classical propositional logic, given here (

http://planetmath.org/DeductionTheoremHoldsForClassicalPropositionalLogic

), in that it uses

induction

on the length of the

deduction

of

B

. In fact, the proof is simpler as only two

axiom schemas

are used:

A

→

(

B

→

A

)

and

(

A

→

B

)

→

(

(

A

→

(

B

→

C

)

)

→

(

A

→

C

)

)

.

Proof.

There are two main cases to consider:

•

If

B

is an axiom or in

Δ

∪

{

A

}

, then

B

,

B

→

(

A

→

B

)

,

A

→

B

is a deduction of

A

→

B

from

Δ

, where

A

→

B

is obtained by

modus ponens

applied to

B

and the axiom

B

→

(

A

→

B

)

. So

Δ

⊢

i

A

→

B

.

•

Now, suppose that

A

1

,

…

,

A

n

is a deduction of

B

from

Δ

∪

{

A

}

, with

B

obtained from earlier

formulas

by modus ponens.

We use induction on the length

n

of deduction of

B

. Note that

n

≥

3

. If

n

=

3

, then

C

and

C

→

B

are either axioms or in

Δ

∪

{

A

}

.

–

If

C

is

A

, then

C

→

B

is either an axiom or in

Δ

. So

Δ

⊢

i

A

→

B

.

–

If

C

→

B

is

A

, then

C

is either an axiom or in

Δ

. Then

ℰ

0

,

C

→

(

A

→

C

)

,

C

,

A

→

C

,

(

A

→

C

)

→

(

(

A

→

(

C

→

B

)

)

→

(

A

→

B

)

)

,

(

A

→

(

C

→

B

)

)

→

(

A

→

B

)

,

A

→

B

is a deduction of

A

→

B

from

Δ

, where

ℰ

0

is a deduction of the

theorem

A

→

A

, followed by an axiom

instance

, then

C

, then the result of modus ponens, then an axiom instance, and finally two applications of modus ponens. Note the second to the last formula is just

(

A

→

A

)

→

(

A

→

B

)

.

–

If

C

and

C

→

B

are axioms or in

Δ

, then

Δ

⊢

i

A

→

B

based on the deduction

C

,

C

→

B

,

B

,

B

→

(

A

→

B

)

,

A

→

B

.

Next, assume there is a deduction

ℰ

of

B

of length

n

>

3

. So one of the earlier formulas is

A

k

→

B

, and a

subsequence

of

ℰ

is a deduction of

A

k

→

B

, which has length less than

n

, and therefore by induction,

Δ

⊢

i

A

→

(

A

k

→

B

)

. Likewise, a subsequence of

ℰ

is a deduction of

A

k

, so by induction,

Δ

⊢

i

A

→

A

k

. With the axiom instance

(

A

→

A

k

)

→

(

(

A

→

(

A

k

→

B

)

)

→

(

A

→

B

)

)

, and two applications of modus ponens, we get

Δ

⊢

i

A

→

B

as required.

In both cases,

Δ

⊢

i

A

→

B

, and the proof is

complete

.
∎

Remark

The deduction theorem can be used to prove the deduction theorem for the first and

second order

intuitionistic predicate logic.

References

1

J. W. Robbin,

Mathematical Logic, A First Course

, Dover Publication (2006)

Title

deduction theorem holds for intuitionistic propositional logic

Canonical

name

DeductionTheoremHoldsForIntuitionisticPropositionalLogic

Date of creation

2013-03-22 19:31:32

Last modified on

2013-03-22 19:31:32

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

21

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 03F55

Classification

msc 03B20

Related topic

AxiomSystemForIntuitionisticLogic

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