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OPERATIONS ON SETS

 
Theorem S.I.1
If x and y are sets then {x,y} = {y,x} and {x,y} != O.

Proof
Let z = {x,y}.
By Definition S.A.1 /\(a) (a:-z <=> (a = x or a = y)).
Thus /\(a) (a:-z <=> (a = y or a = x)). So z = {y,x}.
By Definition S.A.1 x:-{x,y}.
Thus by Axiom ZF1 {x,y} != O.

Definition S.I.2 - Intersection
Let x be a set.
We define that n(x) = {a:-u(x)|/\(b:-x) a:-b}.

Definition S.I.3 - Basic set operations
Let x,y be sets. 
We define that: 
x u y = u({x,y}),
x n y = n({x,y}),
x\y = {a:-x|!(a:-y)}.

x u y, x n y and x\y are sets 
by Axiom ZF5 and Axiom ZF4.

Theorem S.I.4
If x,y,z are sets then the following statements are true.

(1) x u y = z <=>  /\(a) ( a:-z <=> (a:-x or a:-y) ),
(2) x n y = z <=>  /\(a) ( a:-z <=> (a:-x and a:-y) ), 
(3) x\y = z <=>  /\(a) ( a:-z <=> a:-x and !(a:-y) ), 

Proof
(1)
Assume that x u y = z. By Definition S.I.3 z = u({x,y}).
Take any set a.
a:-z <=> a:-u({x,y}) <=> \/(w) w:-{x,y} and a:-w <=> 
\/(w) (w = x or w = y) and a:-w <=> a:-x or a:-y. 
We have showed (=>). 
Assume that /\(a) a:-z <=> (a:-x or a:-y).
Take any set a. 
a:-z <=> (a:-x or a:-y) <=> 
\/(w) (w = x or w = y) and a:-w <=> 
\/(w) w:-{x,y} and a:-w <=> a:-u({x,y}) <=> a:-x u y. 
By Axiom ZF1 z = x u y.
We have showed (<=).
(2)
Notice that:
x n y = n({x,y}) = {a:-u({x,y}) | /\(b:-{x,y}) a:-b} = 
{a:-x u y | a:-x and a:-y}.
Assume that x n y = z. Then z = {a:-x u y | a:-x and a:-y}.
Take any set a.
It is obvious that (a:-z => a:-x and a:-y). 
We have to show (a:-x and a:-y => a:-z). 
If a:-x and a:-y then by Definition S.A.1 
and Definition S.I.3 a:-u({x,y}) = x u y.
Thus by Definition S.A.2 a:-{b:-x u y | b:-x and b:-y}.
So a:-z.
We have showed that 
x n y = z => ( /\(a) a:-z <=> (a:-x and a:-y) ).
Now, assume that /\(a) a:-z <=> a:-x and a:-y.
Take any set a. 
a:-z <=> a:-x and a:-y <=> 
a:-{b:-x u y | b:-x and b:-y} <=> a:- x n y. 
By Axiom ZF1 z =  x n y.
We have showed that 
( /\(a) a:-z <=> (a:-x and a:-y) ) => z = x n y.
(3)
x\y = z <=> z = {b:-x|!(b:-y)} <=> 
(a:-z <=> a:-x and !(a:-y)). 


Theorem S.I.5 - Bool Algebra
If x,y,z are sets then the following statements are true.

(1) x u O = x,
(2) x u y = y u x,
(3) (x u y) u z = x u (y u z),

(4) x n O = O, 
(5) x n y = y n x, 
(6) (x n y) n z = x n (y n z), 

(7) x u (y n z) = (x u y) n (x u z), 
(8) x n (y u z) = (x n y) u (x n z), 

(9) z\(x u y) = (z\x) n (z\y), 
(10) z\(x n y) = (z\x) u (z\y). 

Proof
One can prove the statements above 
applying Theorem S.I.4 and Axiom ZF1.

Definitin S.I.6 - Inclusion
Let x,y be sets. 
We define that x c y <=> ( /\(a) (a:-x => a:-y) ).

Theorem S.I.7
If x,y,z are sets then the following statements are true.

(1) O c x,
(2) x c x, 
(3) x c y and y c z => x c z, 
(4) x c y and y c x => x = y, 
(5) x c y => (z u x) c (z u y), 
(6) x c y => (z n x) c (z n y), 
(7) y\x c y. 

Proof
One can prove the statements above 
applying Theorem S.I.4 and Axiom ZF1.

Theorem S.I.8
If x,y,x are sets then the following statements are true.

(1) x c y u x, 
(2) x c y n x.
(3) y c x => x\(x\y) = y.  

Proof
(1) 
By Theorem S.I.7.1 we have O c y.
By Theorem S.I.7.5 and Theorem S.I.5.1 x = (O u x) c y u x.   
(2) 
By Theorem S.I.7.1 we have O c y.
By Theorem S.I.7.5 and Theorem S.I.5.1 x = (O n x) c y n x. 
(3)
We have 
x\(x\y) = {z:-x | !(z:-x\y)} = {z:-x | !(z:-x and !(x:-y))} = 
{z:-x | !(z:-x) or x:-y} = {z:-x | x:-y} = y.

Theorem S.I.9
If x,z are sets, z:-x and z != O then u(x) != O.

Proof
By Definition S.A.1 /\(a) (a:-u(x) <=> \/(b:-x) a:-b). 
We have z:-x and z != O. Thus we have a:-z. Thus a:-u(x). So u(x) != O.     

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