proof that C_∪ and C_∩ are consequence operators

proof that

C

∪

and

C

∩

are consequence operators

The proof that the

operators

C

∪

and

C

∩

defined in the second
example of

section

3 of the
parent entry (

http://planetmath.org/ConsequenceOperator

) are

consequence operators

is a relatively straightforward matter of checking that they

satisfy

the

defining properties

given there. For convenience, those

definitions

are
reproduced here.

Definition 1

.

Given a set

L

and two

elements

,

X

and

Y

, of this set, the

function

C

∩

⁢

(

X

,

Y

)

:

P

⁢

(

L

)

→

P

⁢

(

L

)

is defined as follows:

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

=

{

X

∪

Z

Y

∩

Z

≠

∅

Z

Y

∩

Z

=

∅

Theorem 1

.

For every choice of two elements,

X

and

Y

, of a given set

L

, the
function

C

∩

⁢

(

X

,

Y

)

is a consequence operator.

Proof.

Property

1:

Since

Z

is a subset of itself and of

X

∪

Z

, it follows that

Z

⊆

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

in either case.

Property 2:

We consider two cases. If

Y

∩

Z

=

∅

, then

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

=

Z

, so

C

∩

⁢

(

X

,

Y

)

⁢

(

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

)

=

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

.

If

Y

∩

Z

≠

∅

, then

Y

∩

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

=

Y

∩

(

X

∪

Z

)

=

(

Y

∩

X

)

∪

(

Y

∩

Z

)

.

Again, since

Y

∩

Z

≠

∅

, we also
have

(

Y

∩

X

)

∪

(

Y

∩

Z

)

≠

∅

, so

C

∩

⁢

(

X

,

Y

)

⁢

(

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

)

=

X

∪

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

=

X

∪

(

X

∪

Z

)

=

X

∪

Z

=

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

So, in both cases, we find that

C

∩

⁢

(

X

,

Y

)

⁢

(

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

)

=

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

.

Property 3:

Suppose that

Z

and

W

are subsets of

L

and that

Z

is a subset
of

W

. Then there are three possibilities:

1.

Y

∩

Z

=

∅

and

Y

∩

W

=

∅

In this case, we have

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

=

Z

and

C

∩

⁢

(

X

,

Y

)

⁢

(

W

)

=

W

, so

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

⊆

C

∩

⁢

(

X

,

Y

)

⁢

(

W

)

.

2.

Y

∩

Z

=

∅

but

Y

∩

W

≠

∅

In this case,

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

=

Z

and

C

∩

⁢

(

X

,

Y

)

⁢

(

W

)

=

X

∪

W

. Since

Z

⊆

W

implies

Z

⊆

X

∪

W

, we have

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

⊆

C

∩

⁢

(

X

,

Y

)

⁢

(

W

)

.

3.

Y

∩

Z

≠

∅

and

Y

∩

W

≠

∅

In this case,

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

=

X

∪

Z

and

C

∩

⁢

(

X

,

Y

)

⁢

(

W

)

=

X

∪

W

. Since

Z

⊆

W

implies

X

∪

Z

⊆

X

∪

W

, we have

C

∩

⁢

(

X

,

Y

)

⁢

(

Z

)

⊆

C

∩

⁢

(

X

,

Y

)

⁢

(

W

)

.

∎

Definition 2

.

Given a set

L

and two elements,

X

and

Y

, of this set, the
function

C

∪

⁢

(

X

,

Y

)

:

P

⁢

(

L

)

→

P

⁢

(

L

)

is defined as follows:

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

=

{

X

∪

Z

Y

∪

Z

=

Z

Z

Y

∪

Z

≠

Z

Theorem 2

.

For every choice of two elements,

X

and

Y

, of a given set

L

, the
function

C

∪

⁢

(

X

,

Y

)

is a consequence operator.

Proof.

Property 1:

Since

Z

is a subset of itself and of

X

∪

Z

, it follows that

Z

⊆

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

in either case.

Property 2:

We consider two cases. If

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

=

Z

, then

C

∪

⁢

(

X

,

Y

)

⁢

(

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

)

=

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

.

If

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

=

X

∪

Z

, then we note that, because

X

∪

(

X

∪

Z

)

=

X

∪

Z

, we must have

C

∪

⁢

(

X

,

Y

)

⁢

(

X

∪

Z

)

=

X

∪

Z

whether or not

Y

∪

(

X

∪

Z

)

=

X

∪

Z

, so

C

∪

⁢

(

X

,

Y

)

⁢

(

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

)

=

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

.

Property 3:

Suppose that

Z

and

W

are subsets of

L

and that

Z

is a subset
of

W

. Then there are three possibilities:

1.

Y

∪

Z

=

Z

and

Y

∪

W

=

W

In this case, we have

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

=

X

∪

Z

and

C

∪

⁢

(

X

,

Y

)

⁢

(

W

)

=

X

∪

W

. Since

Z

⊆

W

implies

X

∪

Z

⊆

X

∪

W

, we have

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

⊆

C

∪

⁢

(

X

,

Y

)

⁢

(

W

)

.

2.

Y

∪

Z

≠

Z

but

Y

∪

W

=

W

In this case,

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

=

Z

and

C

∪

⁢

(

X

,

Y

)

⁢

(

W

)

=

X

∪

W

.
Since

Z

⊆

W

implies

Z

⊆

X

∪

W

, we have

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

⊆

C

∪

⁢

(

X

,

Y

)

⁢

(

W

)

.

3.

Y

∪

Z

≠

Z

and

Y

∪

W

≠

W

In this case,

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

=

Z

and

C

∪

⁢

(

X

,

Y

)

⁢

(

W

)

=

W

,
so

C

∪

⁢

(

X

,

Y

)

⁢

(

Z

)

⊆

C

∪

⁢

(

X

,

Y

)

⁢

(

W

)

.
∎

Title

proof that

C

∪

and

C

∩

are consequence operators

Canonical

name

ProofThatCcupAndCcapAreConsequenceOperators

Date of creation

2013-03-22 16:29:41

Last modified on

2013-03-22 16:29:41

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

21

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 03G25

Classification

msc 03G10

Classification

msc 03B22

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