Part 1: Proof that (a + b) · (à · b) = 0. (a + b). (a. b) = = = = = = (a. b)(a + b) ((ā· b). a) + ((a. b) (5. (a.a) (b ·ā)·a) + ((ā· b) b. (a.a)) +(ā. (5.b)) ·ā)) + (ā· (b · ·5)) (5.0) + (ā.0) b) = 0 + 0 = 0 b b by the commutative law for by the distributive law of over + → by the commutative law for by the distributive law of over + X by the identity law of + by the associative law for by the commutative law for by the complement law for + X X +x +X
Part 1: Proof that (a + b) · (à · b) = 0. (a + b). (a. b) = = = = = = (a. b)(a + b) ((ā· b). a) + ((a. b) (5. (a.a) (b ·ā)·a) + ((ā· b) b. (a.a)) +(ā. (5.b)) ·ā)) + (ā· (b · ·5)) (5.0) + (ā.0) b) = 0 + 0 = 0 b b by the commutative law for by the distributive law of over + → by the commutative law for by the distributive law of over + X by the identity law of + by the associative law for by the commutative law for by the complement law for + X X +x +X
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Associative law for *
Commutative law for *
Complement law for *
Distributive law of * over +
Identity law of +
Universal bound law for *
![Assume that B is a Boolean algebra with operations + and . Prove the following statement.
De Morgan's law for +: For all a and b in B, a + b = a.b.
Proof: Suppose B is a Boolean algebra and a and b are any elements of B. [We must show that a + b = a.b.]
Part 1: Proof that (a + b) · (ā · b) = 0.
(a + b). (ā.b)
(a. b) (a + b)
=
=
=
=
(a.b).a +
((ā. 5). b)
((-a) a) + ((a.). b)
b)
(b. (ā-a)) + (ā- (5.b))
·
b. (a.a)) +(ā· (b.5))
(b 0) + (a. 0)
= 0 + 0
= 0
by the commutative law for
by the distributive law of over +
by the commutative law for.
by the distributive law of over +
by the identity law of +
by the associative law for
by the commutative law for
by the complement law for
+
X
+ X
X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb232dbf3-87c8-40c1-ad6d-cab85c52dec2%2F569a4ad9-f1ec-4d7e-a6a3-5403c9140dbe%2Fay38na9_processed.png&w=3840&q=75)
Transcribed Image Text:Assume that B is a Boolean algebra with operations + and . Prove the following statement.
De Morgan's law for +: For all a and b in B, a + b = a.b.
Proof: Suppose B is a Boolean algebra and a and b are any elements of B. [We must show that a + b = a.b.]
Part 1: Proof that (a + b) · (ā · b) = 0.
(a + b). (ā.b)
(a. b) (a + b)
=
=
=
=
(a.b).a +
((ā. 5). b)
((-a) a) + ((a.). b)
b)
(b. (ā-a)) + (ā- (5.b))
·
b. (a.a)) +(ā· (b.5))
(b 0) + (a. 0)
= 0 + 0
= 0
by the commutative law for
by the distributive law of over +
by the commutative law for.
by the distributive law of over +
by the identity law of +
by the associative law for
by the commutative law for
by the complement law for
+
X
+ X
X
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