deduction theorem holds for classical propositional logic

deduction theorem holds for classical propositional logic

In this entry, we prove that the

deduction theorem holds for classical propositional logic

. For the logic, we use the

axiom system

found in this entry (

http://planetmath.org/AxiomSystemForPropositionalLogic

). To prove the

theorem

, we use the theorem schema

A

→

A

(whose

deduction

can be found here (

http://planetmath.org/AxiomSystemForPropositionalLogic

)).

Proof.

Suppose

A

1

,

…

,

A

n

is a deduction of

B

from

Δ

∪

{

A

}

. We want to find a deduction of

A

→

B

from

Δ

. 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

⊢

A

→

B

.

•

If

B

is obtained from

A

i

and

A

j

by modus ponens. Then

A

j

, say, is

A

i

→

B

. We use

induction

on the length

n

of the deduction of

B

. Note that

n

≥

3

. If

n

=

3

, then the first two

formulas

are

C

and

C

→

B

.

–

If

C

is

A

, then

C

→

B

is either an axiom or in

Δ

. So

A

→

B

, which is just

C

→

B

, is a deduction of

A

→

B

from

Δ

.

–

If

C

→

B

is

A

, then

C

is either an axiom or in

Δ

, and

ℰ

0

,

(

A

→

A

)

→

(

(

A

→

C

)

→

(

A

→

B

)

)

,

C

→

(

A

→

C

)

,

C

,

A

→

C

,

A

→

B

is a deduction of

A

→

B

from

Δ

, where

ℰ

0

is a deduction of

A

→

A

, followed by two axiom

instances

, followed by

C

, followed by results of two applications of modus ponens.

–

If both

C

and

C

→

B

are either axioms are in

Δ

, then

C

,

C

→

B

,

B

,

B

→

(

A

→

B

)

,

A

→

B

is a deduction of

A

→

B

from

Δ

.

Next, assume the deduction

ℰ

of

B

has length

n

>

3

.
A

subsequence

of

ℰ

is a deduction of

A

i

→

B

from

Δ

∪

{

A

}

. This deduction has length less than

n

, so by induction,

Δ

⊢

A

→

(

A

i

→

B

)

,

and therefore by

(

A

→

(

A

i

→

B

)

)

→

(

(

A

→

A

i

)

→

(

A

→

B

)

)

, an axiom instance, and modus ponens,

Δ

⊢

(

A

→

A

i

)

→

(

A

→

B

)

.

Likewise, a subsequence of

ℰ

is a deduction of

A

i

, so by induction,

Δ

⊢

A

→

A

i

. Therefore, an application of modus ponens gives us

Δ

⊢

A

→

B

.

In both cases,

Δ

⊢

A

→

B

and we are done.
∎

We record two corollaries:

Corollary 1

.

(

Proof by Contradiction

). If

Δ

,

A

⊢

⟂

, then

Δ

⊢

¬

⁢

A

.

Proof.

From

Δ

,

A

⊢

⟂

, we have

Δ

⊢

A

→

⟂

by the

deduction theorem

. Since

¬

⁢

A

↔

⟂

, the result follows. ∎

Corollary 2

.

(Proof by Contrapositive). If

Δ

,

A

⊢

¬

⁢

B

, then

Δ

,

B

⊢

¬

⁢

A

.

Proof.

If

Δ

,

A

⊢

¬

⁢

B

, then

Δ

,

A

,

B

⊢

⟂

by the deduction thereom, and therefore

Δ

,

B

⊢

¬

⁢

A

by the deduction theorem again. ∎

Remark

The deduction theorem for the classical

propositional logic

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

second order

predicate logic.

References

1

J. W. Robbin,

Mathematical Logic, A First Course

, Dover Publication (2006)

Title

deduction theorem holds for classical propositional logic

Canonical name

DeductionTheoremHoldsForClassicalPropositionalLogic

Date of creation

2013-03-22 19:13:42

Last modified on

2013-03-22 19:13:42

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

23

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 03F03

Classification

msc 03B99

Classification

msc 03B22

Related topic

AxiomSystemForPropositionalLogic

Defines

proof by contradiction

Defines

proof by contrapositive

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