# Set identity

#### solakis

##### Active member
Given the definition: $$\displaystyle A^{n+1}=A^n\cup A$$ then prove that:

$$\displaystyle (A\cup B)^n =(A\cup B)\cup(A\cap B)$$ for all natural N0s n

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Given the definition: $$\displaystyle A^{n+1}=A^n\cup A$$
What is $A^1$?

then prove that:

$$\displaystyle (A\cup B)^n =(A\cup B)\cup(A\cap B)$$ for all natural N0s n
Hmm, why write $(A\cup B)\cup(A\cap B)$ since $A\cap B$ is included in $A\cup B$?

#### solakis

##### Active member
What is $A^1$?

Hmm, why write $(A\cup B)\cup(A\cap B)$ since $A\cap B$ is included in $A\cup B$?
$$\displaystyle A^1=A$$

Is by definition (or theorem):

1) $$\displaystyle x\in(A\cup B)\Longleftrightarrow x\in A\vee x\in B$$

............................Or.............................

2)$$\displaystyle x\in(A\cup B)\Longleftrightarrow (x\in A\vee x\in B)\vee (x\in A\wedge x\in B)$$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
$$\displaystyle A^1=A$$
Then $A^n=A$ for all $n$.

Is by definition (or theorem):

1) $$\displaystyle x\in(A\cup B)\Longleftrightarrow x\in A\vee x\in B$$

............................Or.............................

2)$$\displaystyle x\in(A\cup B)\Longleftrightarrow (x\in A\vee x\in B)\vee (x\in A\wedge x\in B)$$
Both.

#### solakis

##### Active member
Then $A^n=A$ for all $n$.

Both.
In that case if we put :

xεA=p, xεB= q,how would we prove :

(pvq)vr =pv(qvr)?

Since pvq <=> (pvq)v(p^q)

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
In that case if we put :

xεA=p, xεB= q,how would we prove :

(pvq)vr =pv(qvr)?
Associativity of disjunction holds for all propositions, not necessarily x ∈ A and x ∈ B. The way of proving it depends on axiomatics. In Boolean algebra, for example, this is an axiom.

#### solakis

##### Active member
Associativity of disjunction holds for all propositions, not necessarily x ∈ A and x ∈ B. The way of proving it depends on axiomatics. In Boolean algebra, for example, this is an axiom.
By the way:

In the axioms you suggested in Wikipedia ,two of them can be proved using the other axioms.
Those are the axioms of absorption and the axioms of associativity.

I will prove the axioms of absorption and i will leave it to you to prove the axioms of associativity.

proof:

1) x^(xvy)

2) (xv0)^(xvy)..........using the axiom of identity:av0=a

3) xv(0^y)..............using the axiom of distributivity: av(b^c) = (avb)^(avc)

4) xv(y^0).............using the axiom of commutativity: a^b=b^a

5) xv[(y^0)v0].........using the axiom of identity:av0=a

6) xv[(y^0)v(y^y')]........using the axiom of complements: a^a' = 0 (Note a' is the comlement of a)

7) xv[y^(0vy')]..............using the axiom of distributivity : a^(bvc) = (a^b)v(a^c)

8) xv[y^(y'v0)]..............using the axiom of commutativity: avb=bva

9) xv(y^y') ...................using the axiom of identity: av0=a

10) xv0 .......................using the axiom of complements :y^y'=0

11) x............................using the axiom of identity: av0=a

Also due to the duolity principle we have:

xv(x^y)= x

#### MarkFL

Staff member
...
I will prove the axioms of absorption and i will leave it to you to prove the axioms of associativity....
I'm just curious, why are you leaving work to be done by Evgeny.Makarov?

#### solakis

##### Active member
I'm just curious, why are you leaving work to be done by Evgeny.Makarov?
You mean always or just this time ?

#### MarkFL

Just this one particular instance, the part of your post that I quoted. I was just curious. 