Boolean Algebra: Theorems and Transformations, Study notes of Algebra

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CSE 20 Lecture 9

Boolean Algebra: Theorems and

Transformations

CK Cheng

Theorems & Proofs: 4 Postulates

P1: a+b = b+a, a·b=b·a

(commutative)

P2: a+bc = (a+b)·(a+c) (distributive)

a·(b+c) = a·b + a·c

P3: a+0=a, a·1 = a (identity)

=> a·0=0, a+1=1 (annulment)

P4: a+a’=1, a·a’= 0 (complement)

Theorem 8: For every pair a, b in B a + a’·b = a + b; a·(a’ + b) = a·b Proof: a + a’·b = (a + a’)·(a + b) (P2) = (1)·(a + b) (P4) = a + b (P3)

Theorems and Proofs

Theorem 9: De Morgan’s Law

Theorem: For every pair a, b in set B: (a+b)’ = a’b’, and (ab)’ = a’+b’. Proof: We show that a+b and a’b’ are complementary. In other words, we show that both of the following are true (P4): (a+b)+(a’b’) = 1, (a+b)(a’b’) = 0.

3. Theorems: Switching Algebra vs.

Multiple Valued Boolean Algebra

• Boolean Algebra is termed Switching Algebra when B = {0, 1}
• When |B| > 2, the system is multiple valued.
  • Example: M = {(0, 1, 2, 3), #, &} # 0 1 2 3 0 0 1 2 3 1 1 1 3 3 2 2 3 2 3 3 3 3 3 3

iClicker: M = {(0, 1, 2, 3), #, &} A. Boolean algebra can have only two elements {0, 1}. B. The identity elements are 0 and 3:

• a # 0 = a
• a & 3 = a C. The complement of 1 is 2 D. Two of the above E. None of the above.

Boolean Transform

• Given a Boolean expression, we reduce the expression (#literals, #terms) using laws and theorems of Boolean algebra.
• When B={0,1}, we can use tables to visualize the operation. - The approach follows Shannon’s expansion. - The tables are organized in two dimension space and called Karnaugh maps.

4. Boolean Transformations

Show that a’b’+ab+a’b = a’+b Proof 1: a’b’+ab+a’b = a’b’+(a+a’)b P = a’b’ + b P = a’ + b Theorem 8 Proof 2: a’b’+ab+a’b = a’b’+ab+a’b+a’b Theorem 5 = a’b’ + a’b +ab+a’b P = a’(b’+b) + (a+a’)b P = a’1 +1b P = a’ + b P