Postulate 2	x+0 = x	x⋅1 = x
Postulate 5	x+x′ = 1	x⋅x′ = 0
Theorem 1	x+x = x	x⋅x = x
Theorem 2	x+1 = 1	x⋅0 = 0
Theorem 3, Involution	(x′)′ = x
Postulate 3, commutative	x+y = y+x	xy = yx
Theorem 4, associative	x+(y+z) = (x+y)+z	x(yz) = (xy)z
Postulate 4, distributive	x(y+z) = xy+xz	x+yz = (x+y) (x+z)
Theorem 5, DeMorgan	(x+y)′ = x′y′	(xy)′ = x′+y′
Theorem 6, absorption	x+xy = x	x(x+y) = x

1. A + AB = A

2. A . (A + B) = A

3. A + A'B = A + B

4. A . (A' + B) = AB

5. AB + BC + A'C = AB + A'C

6. (A + B) (B + C) (A' + C) = (A + B) (A' + C)

To find duality convert (+) to (.) or vice-versa

Example: A + B = A . B

Step 1: Find the dual of the function.

Step 2: Complement each literals/varaibles.

Example: F1 = X'YZ' + X'Y'Z

S1 => F1 = (X' + Y + Z') . (X' + Y' + Z)

S2 => F1' = (X + Y' + Z) . (X + Y + Z') //answer

Sum of minterms

A minterm, denoted as m_i, is a product (AND)
uncomplemented X = 1,
complemented X' = 0

Example :
F = x + yz
= x (y + y') (z + z') + (x + x') yz
= xyz + xyz' + xy' z + xy' z' + xyz + x' yz
= x' yz + xy' z' + xy' z + xyz' + xyz
= m₃ + m₄ + m₅ + m₆ + m₇
= ∑(3, 4, 5, 6, 7)

Product of maxterms

A maxterm, denoted as M_i, is a product (OR)
uncomplemented X = 0,
complemented X' = 1

Example :
F = (x+y+z) • (x+y+z') • (x+y'+z) • (x'+y+z)
M₀•M₁•M₂•M₄
= ∏(0, 1, 2, 4)

x	y	z	Minterms	Maxterms	F	F'
0	0	0	m0= x'y'z'	M0= x+y+z	0	1
0	0	1	m1= x'y'z	M1= x+y+z'	1	0
0	1	0	m2= x'yz'	M2= x+y'+z	0	1
0	1	1	m3= x'yz	M3= x+y'+z'	1	0
1	0	0	m4= xy'z'	M4= x'+y+z	1	0
1	0	1	m5= xy'z	M5= x'+y+z'	0	1
1	1	0	m6= xyz'	M6= x'+y'+z	0	1
1	1	1	m7= xyz	M7= x'+y'+z'	0	1

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Canonical form and Standard Form

• Canonical form -
  • Sum of minterms (SOM)
  • Product of maxterms (POM)
• Standard forms (may use less gates) -
  • Sum of products (SOP)
  • Product of sums (POS)
• SOP form may not be in Canonical Form