MHB Proving (A∪B)^n = (A∪B)∪(A∩B) for Natural N0s n

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The discussion centers on proving that (A∪B)^n = (A∪B)∪(A∩B) for all natural numbers n, using the definition A^{n+1}=A^n∪A. Participants clarify that A^1 equals A and explore the implications of union and intersection in set theory. The conversation also touches on the axioms of absorption and associativity in Boolean algebra, noting that the latter is an axiom rather than a theorem. There is a debate about the division of proof responsibilities among participants, highlighting collaborative aspects of mathematical discourse. The thread ultimately emphasizes the importance of foundational axioms in proving set identities.
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Given the definition: $$A^{n+1}=A^n\cup A$$ then prove that:

$$(A\cup B)^n =(A\cup B)\cup(A\cap B)$$ for all natural N0s n
 
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solakis said:
Given the definition: $$A^{n+1}=A^n\cup A$$
What is $A^1$?

solakis said:
then prove that:

$$(A\cup B)^n =(A\cup B)\cup(A\cap B)$$ for all natural N0s n
Hmm, why write $(A\cup B)\cup(A\cap B)$ since $A\cap B$ is included in $A\cup B$?
 
Evgeny.Makarov said:
What is $A^1$?

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

$$A^1=A$$

Is by definition (or theorem):

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

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

2)$$x\in(A\cup B)\Longleftrightarrow (x\in A\vee x\in B)\vee (x\in A\wedge x\in B)$$
 
solakis said:
$$A^1=A$$
Then $A^n=A$ for all $n$.

solakis said:
Is by definition (or theorem):

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

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

2)$$x\in(A\cup B)\Longleftrightarrow (x\in A\vee x\in B)\vee (x\in A\wedge x\in B)$$
Both.
 
Evgeny.Makarov said:
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)
 
solakis said:
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.
 
Evgeny.Makarov said:
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=aAlso due to the duolity principle we have:

xv(x^y)= x
 
solakis said:
...
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?
 
MarkFL said:
I'm just curious, why are you leaving work to be done by Evgeny.Makarov?

You mean always or just this time ?
 
  • #10
solakis said:
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
 
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