substitution theorem for propositional logic

substitution theorem for propositional logic

In this entry, we will prove the

substitution theorem for propositional logic

based on the

axiom system

found here (

http://

planetmath

.org/AxiomSystemForPropositionalLogic

). Besides the

deduction theorem

, below are some additional results we will need to prove the

theorem

:

1.

If

Δ

⊢

A

→

B

and

Γ

⊢

B

→

C

, then

Δ

,

Γ

⊢

A

→

C

.

2.

Δ

⊢

A

and

Δ

⊢

B

iff

Δ

⊢

A

∧

B

.

3.

⊢

A

↔

A

.

4.

⊢

A

↔

¬

¬

A

(

law of double negation

).

5.

⟂

→

A

(ex falso quodlibet)

6.

Δ

⊢

A

implies

Δ

⊢

B

iff

Δ

⊢

A

→

B

.

The proofs of these results can be found here (

http://planetmath.org/SomeTheoremSchemasOfPropositionalLogic

).

Theorem 1

.

(Substitution Theorem) Suppose

p

1

,

…

,

p

m

are all the

propositional variables

, not necessarily distinct, that

occur in

order in

A

, and if

B

1

,

…

,

B

m

,

C

1

,

…

,

C

m

are wff’s such that

⊢

B

i

↔

C

i

, then

⊢

A

[

B

1

/

p

1

,

…

,

B

m

/

p

m

]

↔

A

[

C

1

/

p

1

,

…

,

C

m

/

p

m

]

where

A

⁢

[

X

1

/

p

1

,

…

,

X

m

/

p

m

]

is the wff obtained from

A

by replacing

p

i

by the wff

X

i

via

simultaneous substitution

.

Proof.

We do

induction

on the number

n

of

→

in wff

A

.

If

n

=

0

,

A

is either a propositional variable, say

p

, or

⟂

, which respectively means that

A

⁢

[

B

/

p

]

↔

A

⁢

[

C

/

p

]

is either

B

↔

C

or

⟂

⁣

↔

⁣

⟂

. The former is the

assumption

and the latter is a theorem.

Suppose now

A

has

n

+

1

occurrences of

→

. We may write

A

as

X

→

Y

uniquely by

unique readability

. Also, both

X

and

Y

have at most

n

occurrences of

→

.

Let

A

1

be

A

⁢

[

B

1

/

p

1

,

…

,

B

m

/

p

m

]

and

A

2

be

A

⁢

[

C

1

/

p

1

,

…

,

C

m

/

p

m

]

. Then

A

1

is

X

1

→

Y

1

and

X

2

→

Y

2

, where

X

1

is

X

⁢

[

B

1

/

p

1

,

…

,

B

k

/

p

k

]

,

Y

1

is

Y

⁢

[

B

k

+

1

/

p

k

+

1

,

…

,

B

m

/

p

m

]

,

X

2

is

X

⁢

[

C

1

/

p

1

,

…

,

C

k

/

p

k

]

, and

Y

2

is

Y

⁢

[

C

k

+

1

/

p

k

+

1

,

…

,

C

m

/

p

m

]

.

Then

by induction

⊢

X

1

↔

X

2

(1)

by 2 above

⊢

X

1

→

X

2

and

⊢

X

2

→

X

1

(2)

by induction

⊢

Y

1

↔

Y

2

(3)

by 2 above

⊢

Y

1

→

Y

2

and

⊢

Y

2

→

Y

1

(4)

since

⁢

A

1

⁢

is

⁢

X

1

→

Y

1

A

1

⊢

X

1

→

Y

1

(5)

by applying 1 to

⊢

X

2

→

X

1

⁢

and

⁢

(

5

)

A

1

⊢

X

2

→

Y

1

(6)

by applying 1 to

⁢

(

6

)

⁢

and

⊢

Y

1

→

Y

2

A

1

⊢

X

2

→

Y

2

(7)

by the deduction theorem

⊢

A

1

→

A

2

(8)

by a

similar

reasoning as above

⊢

A

2

→

A

1

(9)

by applying 2 to

⁢

(

8

)

⁢

and

⁢

(

9

)

⊢

A

1

↔

A

2

(10)

∎

As a corollary, we have

Corollary 1

.

If

⊢

B

↔

C

, then

⊢

A

[

B

/

s

(

p

)

]

↔

A

[

C

/

s

(

p

)

]

, where

p

is a propositional variable that occurs in

A

,

s

⁢

(

p

)

is a set of positions of occurrences of

p

in

A

, and the wff

A

⁢

[

X

/

s

⁢

(

p

)

]

is obtained by replacing all

p

that occur in the positions in

s

⁢

(

p

)

in

A

by wff

X

.

Proof.

For any propositional variable

q

not being replaced, use the corresponding theorem

⊢

q

↔

q

, and then apply the substitution theorem. ∎

Remark

. What about

⊢

B

[

A

/

p

]

↔

C

[

A

/

p

]

, given

⊢

B

↔

C

? Here,

B

⁢

[

A

/

p

]

and

C

⁢

[

A

/

p

]

are wff’s obtained by uniform substitution of

p

(all occurrences of

p

) in

B

and

C

respectively. Since

B

⁢

[

A

/

p

]

↔

C

⁢

[

A

/

p

]

is just

(

B

↔

C

)

[

A

/

p

]

, an instance of the schema

B

↔

C

by assumption, the result follows directly if we assume

B

↔

C

is a theorem schema.

Using the substitution theorem, we can easily derive more theorem schemas, such as

7.

(

A

→

B

)

↔

(

¬

B

→

¬

A

)

(Law of Contraposition)

8.

A

→

(

¬

B

→

¬

(

A

→

B

)

)

9.

(

(

A

→

B

)

→

A

)

→

A

(Peirce’s Law)

Proof.

7.

Since

(

¬

B

→

¬

A

)

→

(

A

→

B

)

is already a theorem schema, we only need to show

⊢

(

A

→

B

)

→

(

¬

B

→

¬

A

)

. By law of double negation (4 above) and the substitution theorem, it is enough to show that

⊢

(

¬

¬

A

→

¬

¬

B

)

→

(

¬

B

→

¬

A

)

. But this is just an instance of an

axiom schema

. Combining the two schemas, we get

⊢

(

A

→

B

)

↔

(

¬

B

→

¬

A

)

.

8.

First, observe that

A

,

A

→

B

⊢

B

by

modus ponens

. Since

⊢

B

↔

¬

¬

B

, we have

A

,

A

→

B

⊢

¬

⁢

¬

⁢

B

by the substitution theorem. So

A

,

A

→

B

,

¬

B

⊢

⟂

by the deduction theorem, and

A

,

¬

B

⊢

(

A

→

B

)

→

⟂

by the deduction theorem again. Apply the

deduction

two more times, we get

⊢

A

→

(

¬

B

→

¬

(

A

→

B

)

)

.

9.

To show

⊢

(

(

A

→

B

)

→

A

)

→

A

, it is enough to show

⊢

¬

A

→

¬

(

(

A

→

B

)

→

A

)

by

7

and modus ponens, or

¬

A

⊢

¬

(

(

A

→

B

)

→

A

)

by the deduction theorem. Now, since

⊢

X

∧

Y

↔

¬

(

X

→

¬

Y

)

(as they are the same thing, and because

C

↔

C

is a theorem schema), by the law of double negation and the substitution theorem,

⊢

X

∧

¬

Y

↔

¬

(

X

→

Y

)

, and we have

⊢

(

A

→

B

)

∧

¬

A

↔

¬

(

(

A

→

B

)

→

A

)

. So to show

¬

A

⊢

¬

(

(

A

→

B

)

→

A

)

, it is enough to show

¬

A

⊢

(

A

→

B

)

∧

¬

A

, which is enough to show that

¬

⁢

A

⊢

A

→

B

and

¬

⁢

A

⊢

¬

⁢

A

, according to a

meta-theorem

found here (

http://planetmath.org/SomeTheoremSchemasOfPropositionalLogic

). To show

¬

⁢

A

⊢

A

→

B

, it is enough to show

{

¬

⁢

A

,

A

}

⊢

B

, and

A

,

¬

A

,

⟂

,

⟂

→

B

,

B

is such a deduction. The second statement

¬

⁢

A

⊢

¬

⁢

A

is clear.

∎

As an application, we prove the following useful meta-theorems of

propositional logic

:

Proposition 1

.

There is a wff

A

such that

Δ

⊢

A

and

Δ

⊢

¬

⁢

A

iff

Δ

⊢

⟂

Proof.

Assume the former. Let

ℰ

1

be a deduction of

A

from

Δ

and

ℰ

2

a deduction of

¬

⁢

A

from

Δ

, then

ℰ

1

,

ℰ

2

,

⟂

is a deduction of

⟂

from

Δ

.

Conversely

, assume the later. Pick any wff

A

(if necessary, pick

⟂

). Then

⟂

→

A

by ex falso quodlibet. By modus ponens, we have

Δ

⊢

A

. Similarly,

Δ

⊢

¬

⁢

A

.
∎

Proposition 2

.

If

Δ

,

A

⊢

B

and

Δ

,

¬

⁢

A

⊢

B

, then

Δ

⊢

B

Proof.

By assumption, we have

Δ

⊢

A

→

B

and

Δ

⊢

¬

⁢

A

→

B

. Using modus ponens and the theorem schema

(

A

→

B

)

→

(

¬

B

→

¬

A

)

, we have

Δ

⊢

¬

⁢

B

→

¬

⁢

A

, or

Δ

,

¬

⁢

B

⊢

¬

⁢

A

.

Similarly,

Δ

⊢

¬

⁢

B

→

¬

⁢

¬

⁢

A

. By the law of double negation and the substitution theorem, we have

Δ

⊢

¬

⁢

B

→

A

, or

Δ

,

¬

⁢

B

⊢

A

.

By the previous

proposition

,

Δ

,

¬

⁢

B

⊢

⟂

, or

Δ

⊢

¬

⁢

¬

⁢

B

. Applying the substitution theorem and the law of double negation, we have

Δ

⊢

B

.

∎

Title

substitution theorem for propositional logic

Canonical name

SubstitutionTheoremForPropositionalLogic

Date of creation

2013-03-22 19:33:08

Last modified on

2013-03-22 19:33:08

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

17

Author

CWoo (3771)

Entry type

Result

Classification

msc 03B05

Related topic

AxiomSystemForPropositionalLogic

Defines

substitution theorem

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