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Set identity

solakis

Active member
Dec 9, 2012
322
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
Jan 30, 2012
2,502
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
Dec 9, 2012
322
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
Jan 30, 2012
2,502
\(\displaystyle A^1=A\)
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
Dec 9, 2012
322

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
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
Dec 9, 2012
322
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

Administrator
Staff member
Feb 24, 2012
13,775
...
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
Dec 9, 2012
322

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You mean always or just this time ?
Just this one particular instance, the part of your post that I quoted. I was just curious. :D