properties of substitutability

properties of substitutability

In this entry, we list some basic

properties of substitutability

in

first order logic

with respect to

commutativity

.

Proposition 1

.

If

x

,

y

are distinct variables, then

t

⁢

[

s

/

x

]

⁢

[

r

/

y

]

=

t

⁢

[

r

/

y

]

⁢

[

s

⁢

[

r

/

y

]

/

x

]

,

provided that

x

does not

occur in

r

.

Proof.

Suppose

x

and

y

are distinct variables, and

r

,

s

,

t

are terms. We do

induction

on the

complexity

of

t

.

1.

First, suppose

t

is a

constant symbol

. Then LHS and RHS are both

t

.

2.

Next, suppose

t

is a variable.

–

If

t

is

x

, then LHS

=

s

⁢

[

r

/

y

]

, and since

y

is not

x

, RHS

=

x

⁢

[

s

⁢

[

r

/

y

]

/

x

]

=

s

⁢

[

r

/

y

]

.

–

If

t

is

y

, then LHS

=

y

⁢

[

r

/

y

]

=

r

, since

x

is not

y

, and RHS

=

r

⁢

[

s

⁢

[

r

/

y

]

/

x

]

=

r

, since

x

does not occur in

r

.

–

If

t

is neither

x

nor

y

, then both sides are

t

.

In all three cases, LHS

=

RHS.

3.

Finally, suppose

t

is of the form

f

⁢

(

t

1

,

…

,

t

n

)

. Then LHS

=

f

⁢

(

t

1

⁢

[

s

/

x

]

,

…

,

t

n

⁢

[

s

/

x

]

)

⁢

[

r

/

y

]

=

f

⁢

(

t

1

⁢

[

s

/

x

]

⁢

[

r

/

y

]

,

…

,

t

n

⁢

[

s

/

x

]

⁢

[

r

/

y

]

)

, which, by induction, is

f

⁢

(

t

1

⁢

[

r

/

y

]

⁢

[

s

⁢

[

r

/

y

]

/

x

]

,

…

,

t

n

⁢

[

r

/

y

]

⁢

[

s

⁢

[

r

/

y

]

/

x

]

)

or

f

⁢

(

t

1

⁢

[

r

/

y

]

,

…

,

t

n

⁢

[

r

/

y

]

)

⁢

[

s

⁢

[

r

/

y

]

/

x

]

=

f

⁢

(

t

1

,

…

,

t

n

)

⁢

[

r

/

y

]

⁢

[

s

⁢

[

r

/

y

]

/

x

]

=

RHS.

∎

Now, if

s

is

y

, then

t

⁢

[

y

/

x

]

⁢

[

r

/

y

]

=

t

⁢

[

r

/

y

]

⁢

[

r

/

x

]

, and we record the following corollary:

Corollary 1

.

If

x

is not in

r

and

y

not in

t

, then

t

⁢

[

y

/

x

]

⁢

[

r

/

y

]

=

t

⁢

[

r

/

x

]

.

The only thing we need to show is the case when

x

is

y

, but this is also clear, as

t

⁢

[

y

/

x

]

⁢

[

r

/

y

]

=

t

⁢

[

x

/

x

]

⁢

[

r

/

x

]

=

t

⁢

[

r

/

x

]

.

With respect to

formulas

, we have a

similar

proposition

:

Proposition 2

.

If

x

,

y

are distinct variables, then

A

⁢

[

t

/

x

]

⁢

[

s

/

y

]

=

A

⁢

[

s

/

y

]

⁢

[

t

⁢

[

s

/

y

]

/

x

]

,

provided that

x

does not occur in

s

, and

t

and

s

are respectively

free for

x

and

y

in

A

.

Proof.

Suppose

x

and

y

are distinct variables,

s

,

t

terms, and

A

a wff. We do induction on the complexity of

A

.

1.

First, suppose

A

is atomic.

–

A

is of the form

t

1

=

t

2

, then LHS is

t

1

⁢

[

t

/

x

]

⁢

[

s

/

y

]

=

t

2

⁢

[

t

/

x

]

⁢

[

s

/

y

]

and we can apply the previous

equation

to both

t

1

and

t

2

to get RHS.

–

If

A

is of the form

R

⁢

(

t

1

,

…

,

t

n

)

, then LHS is

R

⁢

(

t

1

⁢

[

t

/

x

]

⁢

[

s

/

y

]

,

…

,

t

n

⁢

[

t

/

x

]

⁢

[

s

/

y

]

)

, and we again apply the previous equation to each

t

i

to get RHS.

2.

Next, suppose

A

is of the form

B

→

C

. Then LHS

=

B

⁢

[

t

/

x

]

⁢

[

s

/

y

]

→

C

⁢

[

t

/

x

]

⁢

[

s

/

y

]

, and, by induction, is

B

⁢

[

s

/

y

]

⁢

[

t

⁢

[

s

/

y

]

/

x

]

→

C

⁢

[

s

/

y

]

⁢

[

t

⁢

[

s

/

y

]

/

x

]

=

RHS.

3.

Finally, suppose

A

is of the form

∃

z

⁢

B

.

–

x

is

z

. Then

A

⁢

[

t

/

x

]

⁢

[

s

/

y

]

=

A

⁢

[

s

/

y

]

, and

A

⁢

[

s

/

y

]

⁢

[

t

⁢

[

s

/

y

]

/

x

]

=

A

⁢

[

s

/

y

]

since

x

is bound in

A

⁢

[

s

/

y

]

.

–

x

is not

z

. Then

A

⁢

[

t

/

x

]

⁢

[

s

/

y

]

=

(

∃

z

⁢

B

⁢

[

t

/

x

]

)

⁢

[

s

/

y

]

.

*

y

is

z

. Then

(

∃

z

⁢

B

⁢

[

t

/

x

]

)

⁢

[

s

/

y

]

=

∃

z

⁢

B

⁢

[

t

/

x

]

. On the other hand,

A

⁢

[

s

/

y

]

⁢

[

t

⁢

[

s

/

y

]

/

x

]

=

A

⁢

[

t

⁢

[

s

/

y

]

/

x

]

=

∃

z

⁢

B

⁢

[

t

⁢

[

s

/

y

]

/

x

]

. By induction,

t

,

s

are free for

x

,

y

in

B

, and

B

⁢

[

t

/

x

]

=

B

⁢

[

t

⁢

[

s

/

y

]

/

x

]

, the result follows.

*

y

is not

z

. Then

(

∃

z

⁢

B

⁢

[

t

/

x

]

)

⁢

[

s

/

y

]

=

∃

z

⁢

B

⁢

[

t

/

x

]

⁢

[

s

/

y

]

. On the other hand,

A

⁢

[

s

/

y

]

⁢

[

t

⁢

[

s

/

y

]

/

x

]

=

(

∃

z

⁢

B

⁢

[

s

/

y

]

)

⁢

[

t

⁢

[

s

/

y

]

/

x

]

=

∃

z

⁢

B

⁢

[

s

/

y

]

⁢

[

t

⁢

[

s

/

y

]

/

x

]

since

x

is not

z

. By induction again,

t

,

s

are free for

x

,

y

in

B

, and

B

⁢

[

t

/

x

]

=

B

⁢

[

t

⁢

[

s

/

y

]

/

x

]

, the result follows once more.

∎

Now, if

t

is

y

, then

A

⁢

[

y

/

x

]

⁢

[

s

/

y

]

=

A

⁢

[

s

/

y

]

⁢

[

s

/

x

]

, and we record the following corollary:

Corollary 2

.

If

y

is not free in

A

, and is free for

x

in

A

, then

A

⁢

[

y

/

x

]

⁢

[

s

/

y

]

=

A

⁢

[

s

/

x

]

.

Title

properties of substitutability

Canonical name

PropertiesOfSubstitutability

Date of creation

2013-03-22 19:35:51

Last modified on

2013-03-22 19:35:51

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

23

Author

CWoo (3771)

Entry type

Result

Classification

msc 03B10

Classification

msc 03B05

\@unrecurse

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