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- Thread starter solakis
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- Jan 30, 2012

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What is $A^1$?Given the definition: \(\displaystyle A^{n+1}=A^n\cup A\)

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

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

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\(\displaystyle A^1=A\)What is $A^1$?

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

Is by definition (or theorem):

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

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

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

- Jan 30, 2012

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Then $A^n=A$ for all $n$.\(\displaystyle A^1=A\)

Both.Is by definition (or theorem):

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

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

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

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In that case if we put :Then $A^n=A$ for all $n$.

Both.

xεA=p, xεB= q,how would we prove :

(pvq)vr =pv(qvr)?

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

- Jan 30, 2012

- 2,502

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.In that case if we put :

xεA=p, xεB= q,how would we prove :

(pvq)vr =pv(qvr)?

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- #7

By the way: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.

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

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

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- #8

I'm just curious, why are you leaving work to be done by...

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

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- #9

You mean always or just this time ?I'm just curious, why are you leaving work to be done byEvgeny.Makarov?

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- #10

Just this one particular instance, the part of your post that I quoted. I was just curious.You mean always or just this time ?