Q#1.
LetΒ xβAβ©C
Case 1,
SinceΒ AβB,xβAβΉxβB
xβBβΉxβBβ©C
Case 2,
xβC
xβCβΉxβBβ©CΒ sinceΒ xβB
Thus, in either cases,Β xβAβ©CβΉxβBβ©C.
So,
βx(xβAβ©CβΉxβBβ©C)
We now have that,
Aβ©CβBβ©CΒ β .
Q#2.
LetΒ (x,y)β(Aβ©B)Γ(Cβ©D)
βΊxβ(Aβ©B)Γ(Cβ©D)Γy
βΊxβAβ©Bβ§yβCβ©D
βΊxβAβ§xβBβ§yβCβ§yβD
βΊx(AΓC)yβ§x(BΓD)y
βΊx(AΓC)β©(BΓD)y
Therefore,
(Aβ©B)Γ(Cβ©D)=(AΓC)β©(BΓD)Β as requiredΒ β Β .
Q#3.
To begin with, we will first show thatΒ AΓ(Bβ©C)β(AΓB)β©(AΓC). SupposeΒ (x,y)βAΓ(Bβ©C). This means that,Β xβAΒ andΒ yβBβ©CΒ by definition of Cartesian product. By definition of intersection, it follows thatΒ yβBΒ andΒ yβC.SinceΒ xβAΒ andΒ yβB, it follows thatΒ (x,y)βAΓBΒ by definition of Cartesian product. Also, sinceΒ xβAΒ andΒ yβCΒ it follows thatΒ (x,y)βAΓCΒ by definition ofΒ Γ. Now, we haveΒ (x,y)βAΓBΒ andΒ (x,y)βAΓCΒ and so,Β (x,y)β(AΓB)β©(AΓC). Having shown thatΒ (x,y)βAΓ(Bβ©C)βΉ(x,y)β(AΓB)β©(AΓC), we have,Β AΓ(Bβ©C)β(AΓB)β©(AΓC).
Next, we show thatΒ (AΓB)β©(AΓC)βAΓ(Bβ©C),
SupposeΒ (x,y)β(AΓB)β©(AΓC). By definition of intersection, it means that,Β (x,y)βAΓBΒ andΒ (x,y)βAΓC. By definition of Cartesian product,Β (x,y)βAΓBΒ meansΒ xβAΒ andΒ yβBΒ also,Β (x,y)βAΓCΒ meansΒ xβAΒ andΒ yβC. We have that,Β yβBΒ andΒ yβCΒ so,Β yβBβ©CΒ by definition of intersection. Thus, we have deduced thatΒ xβAΒ andΒ yβBβ©C, soΒ (x,y)βAΓ(Bβ©C). Hence,
(AΓB)β©(AΓC)βAΓ(Bβ©C).
In summary, we have shown thatΒ AΓ(Bβ©C)β(AΓB)β©(AΓC)Β andΒ (AΓB)β©(AΓC)βAΓ(Bβ©C). It follows that,Β AΓ(Bβ©C)=(AΓB)β©(AΓC)Β as requiredβ .
Q#4.
SupposeΒ xβp(A)βͺp(B),
Then,Β xβp(A)Β orΒ xβp(B)
Hence,
xβAΒ orΒ xβB
We need to show thatΒ xβp(AβͺB),Β that is, we need to show thatΒ xβAβͺB
SupposeΒ yβxΒ then,Β yβAΒ orΒ yβBΒ sinceΒ xβAΒ orΒ xβB
Thus,Β yβxβΉyβ(AβͺB)Β and so,Β xβ(AβͺB).
Hence,
xβp(AβͺB)
β΄xβp(A)βͺp(B)βΉxβp(AβͺB)
and so,
p(A)βͺp(B)βp(AβͺB)Β as requiredΒ β Β .
Q#5.
Suppose thatΒ AΒ andΒ BΒ are sets andΒ p(A)βp)(B)
By definition of a power set,Β Aβp(A).SinceΒ Aβp(A)Β andΒ p(A)βp(B), we know thatΒ Aβp(B)Β by definition of subset.
So, by definition of power set,Β AβBΒ β .