Proof.

We will prove only the rst as the second is dual.

By denition of joins, it is enough to prove

y

w

F S

S

,

y

w

F

f

F

X

j

X

2

S

g

.

Really,

y

w

F S

S

, 8

x

2

S

S

:

y

w

x

, 8

X

2

S

8

x

2

X

:

y

w

x

, 8

X

2

S

:

y

w

F

X

,

y

w

F

f

F

X

j

X

2

S

g

.

2.1.6 Distributivity of lattices

Denition 2.53.

A

distributive

lattice is such lattice

A

that for every

x; y; z

2

A

1.

x

u

(

y

t

z

) = (

x

u

y

)

t

(

x

u

z

)

;

2.

x

t

(

y

u

z

) = (

x

t

y

)

u

(

x

t

z

)

.

Theorem 2.54.

For a lattice to be distributive it is enough just one of the conditions:

1.

x

u

(

y

t

z

) = (

x

u

y

)

t

(

x

u

z

)

;

2.

x

t

(

y

u

z

) = (

x

t

y

)

u

(

x

t

z

)

.

Proof.

(

x

t

y

)

u

(

x

t

z

) = ((

x

t

y

)

u

x

)

t

((

x

t

y

)

u

z

) =

x

t

((

x

u

z

)

t

(

y

u

z

)) = (

x

t

(

x

u

z

))

t

(

y

u

z

) =

x

t

(

y

u

z

)

(applied

x

u

(

y

t

z

) = (

x

u

y

)

t

(

x

u

z

)

twice).

2.1.7 Dierence and complement

Denition 2.55.

Let

A

be a distributive lattice with least element

0

. The

dierence

(denoted

a

n

b

) of elements

a

and

b

is such

c

2

A

that

b

u

c

= 0

and

a

t

b

=

b

t

c

. I will call

b

substractive

from

a

when

a

n

b

exists.

Theorem 2.56.

If

A

is a distributive lattice with least element

0

, there exists no more than one

dierence of elements

a

,

b

.

Proof.

Let

c

and

d

be both dierences

a

n

b

. Then

b

u

c

=

b

u

d

= 0

and

a

t

b

=

b

t

c

=

b

t

d

. So

c

=

c

u

(

b

t

c

) =

c

u

(

b

t

d

) = (

c

u

b

)

t

(

c

u

d

) = 0

t

(

c

u

d

) =

c

u

d:

Similarly

d

=

d

u

c

. Consequently

c

=

c

u

d

=

d

u

c

=

d

.

Denition 2.57.

I will call

b

complementive

to

a

i there exists

c

2

A

such that

b

u

c

= 0

and

b

t

c

=

a

.

Proposition 2.58.

b

is complementive to

a

i

b

is substractive from

a

and

b

v

a

.

Proof.

(

.

Obvious.

)

.

We deduce

b

v

a

from

b

t

c

=

a

. Thus

a

t

b

=

a

=

b

t

c

.

Proposition 2.59.

If

b

is complementive to

a

then

(

a

n

b

)

t

b

=

a

.

Proof.

Because

b

v

a

by the previous proposition.

Denition 2.60.

Let

A

be a bounded distributive lattice. The

complement

(denoted

a

) of an

element

a

2

A

is such

b

2

A

that

a

u

b

= 0

and

a

t

b

= 1

.

Proposition 2.61.

If

A

is a bounded distributive lattice then

a

= 1

n

a

.

Proof.

b

=

a

,

b

u

a

= 0

^

b

t

a

= 1

,

b

u

a

= 0

^

1

t

a

=

a

t

b

,

b

= 1

n

a

.

Corollary 2.62.

If

A

is a bounded distributive lattice then exists no more than one complement

of an element

a

2

A

.

2.1 Order theory

21