This preview shows pages 1–3. Sign up to view the full content.
ECE 15A
Homework 2 Solutions
1. Write out the proof that
a(a+b) = a
in theorem 4, referring each step to
the correct postulate:
a(a+b) = a
aa + ab
by Distributive Law
a + ab by Theorem 2
a1+ab
P2
a(1+b) by Distributive Law
a=1a
T2
2. Prove that in every Boolean Algebra every triple of elements a,b,c
satisfies the identity ab+bc+ca=(a+b)(b+c)(c+a)
ab+bc+ca
b(a+c)+ca
Distributive Law
(b+ac)(a+c+ac)
Distributive Law
(b+a)(b+c)(a+a+c)(a+c+c) Distributive Law
(b+a)(b+c)(a+c)
3. Prove, that if
a + x
= b + x
and
a + x’
= b + x’
, then
a = b
.
a = a + ax + ax’
Thm 4
b = b + bx + bx’
Thm 4
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documenta =
a + ax + ax’
=
aa + ax + ax’ + xx’
P2, P4
=
a ( a + x) + x’( a + x)
P3
=
( a + x ) ( a + x’)
P3
= ( b + x ) ( b + x’)
Given
= bb + bx’ + xb + xx’
P3
= b
+ bx + bx’
P4
= b
4.
Prove, that if
ax = bx
and
ax’ = bx’
, then
a = b
.
a = a ( x + x’)
P4
=
a
x
+
a
x
’
P3
=
b
x
+
b
x
’
Given
=
b
(
x
+
x
’
)
P3
=
b
P
4
==> a = b
5. Show that set {a,b,c,d} with operations (+) and (.) as defined is a
Boolean algebra.
To prove that the operations are a Boolean algebra, we have to
show that the postulates P1 P4 (see lecture notes #3, slide 14) hold.
This is the end of the preview. Sign up
to
access the rest of the document.
 Winter '08
 M

Click to edit the document details