There is a couple answers to this already, which have helped me get this far. But I have a few questions left.
- Is there a rule in Set Theory that says I have to assign the list to a "Set" before I can use the notation for sets. Easier if I show what I mean,
Can I say {a}∪∅={a}, or do I have to assign Q={a}, and Q∪∅=Q
- Is this sufficient proof for the question:
Let A = {a, b} and the power set P(A) = {∅,{a},{b},A}.
We consider the operations + to be ∪, · to be ∩, and complement to be set complement. Consider 1 to be A and 0 to be ∅.
My answer seems... bulky. But I with the help of previous answers, I believe I'm on track. My proof is below:
If this defines a Boolean algebra, it must have the axioms that govern these operations:
Identity – There exists two elements, 0 and 1, in A such that for every x ∈ P(A),
x + 0 = x and x ∙ 1 = x
Substituting x for elements of P(A),1=A,0=∅ , and set operations we have:
{a} ∪ ∅={a},{b} ∪ ∅={b},A ∪ ∅=A,∅ ∪ ∅=∅
{a} ∩ A={a},{b} ∩ A={b},A ∩ A=A,∅ ∩ A=∅
Complement – For every x ∈P(A), there exists a unique( x ) ̅∈P(A) such that,
x + x ̅ = 1 and x ∙ x ̅ = 0
Substituting x for elements of P(A),1=A,0=∅ , and set operations we have:
{a}∪{∅,{b},A}=A,{b}∪{∅,{a},A}=A,A∪∅=A,∅∪A=A
{a}∩{∅,{b},A}=∅,{b}∩{∅,{a},A}=∅,A∩∅=∅,∅∩A=∅
Commutativity – For every x,y ∈ P(A),
x + y = y+x and x ∙ y = y ∙ x
Substituting x for elements of P(A),1=A,0=∅ , and set operations we have:
{a}∪{b}={b}∪{a},{a}∪A=A∪{a},{a}∪∅=∅∪{a},{b}∪A=A∪{b},{b}∪∅=∅∪{b},A∪∅=∅∪A
a{a}∩{b}={b}∩{a},{a}∩A=A∩{a},{a}∩∅=∅∩{a},{b}∩A=A∩{b},{b}∩∅=∅∩{b},A∩∅=∅∩A
Distributivity – For every x,yz ∈ P(A),
x ∙(y+z)= x∙y+x∙z and x+y ∙ z =(x+y) ∙(x+z)
Substituting x for elements of P(A),1=A,0=∅ , and set operations we have:
{a}∩({b}∪A)=({a}∩{b})∪({a}∩A),{a}∩({b}∪∅)=({a}∩{b})∪({a}∩∅),
{a}∩(A∪∅)=({a}∩A)∪({a}∩∅), {b}∩({a}∪A)=({b}∩{a})∪({b}∩A),
{b}∩({a}∪∅)=({b}∩{a})∪({b}∩∅),{b}∩(A∪∅)=({b}∩A)∪({b}∩∅),
A ∩({a}∪{b})=(A∩{a})∪(A∩{b}),A∩({a}∪∅)=(A∩{a})∪(A∩∅),
A∩({b}∪∅)=(A∩{b})∪(A∩∅),∅ ∩({a}∪{b})=(∅∩{a})∪(∅∩{b}),
∅ ∩({a}∪A)=(∅∩{a})∪(∅∩A),∅ ∩({b}∪A)=(∅∩{b})∪(∅∩A)
{a}∪({b}∩A)=({a}∪{b})∩({a}∪A),{a}∪({b}∩∅)=({a}∪{b})∩({a}∪∅),
{a}∪(A∩∅)=({a}∪A)∩({a}∪∅), {b}∪({a}∩A)=({b}∪{a})∩({b}∪A),
{b}∪({a}∩∅)=({b}∪{a})∩({b}∪∅),{b}∪(A∩∅)=({b}∪A)∩({b}∪∅),
A∪({a}∩{b})=(A∪{a})∩(A∪{b}),A∪({a}∩∅)=(A∪{a})∩(A∪∅),
A∪({b}∩∅)=(A∪{b})∩(A∪∅),∅∪({a}∩{b})=(∅∪{a})∩(∅∪{b}),
∅∪({a}∩A)=(∅∪{a})∩(∅∪A),∅∪({b}∩A)=(∅∪{b})∩(∅∪A)
As you can see, my proof is very bulky. But I at least believe I have the concept down. I am a little iffy on the complement and distributive axiom though.