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| Principle of Duality |
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| The dual of any statement in a Boolean algebra is the statement obtained by interchanging + and ., and simultaneously inter-changing the elements 0 and 1 in the statement. |
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| Example: |
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| (i) The dual of the statement (x . 1 )(0 + x') = 0 is given by |
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| (x + 0)(1 . x') = 1 |
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| (ii) The dual of the statement |
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| (x + y)(y + z) = x.z + y is |
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| x.y + y.z = (x + z).y |
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| Note: |
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| The duality principle has many applications. If the dual of an algebraic expression is desired, we simply interchange + and . and replace 1 by 0 and 0 by 1. |
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Let B be a Boolean algebra. Let x, y, z B. |
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| Theorem 1 (a) (Idempotent Law): |
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| x + x = (x + x) . 1 by Axiom 2(b) |
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| (Refer to Axion in previous topic) |
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| = (x + x) . (x + x') Axiom 5(a) |
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| = x + x.x' Axiom 4(b) |
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| = x + 0 Axiom 5(b) |
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| = x Axiom 2(a) |
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| Theorem 1(b) (Idempotent Law): |
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| x.x = x |
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| x.x = x.x + 0 2(a) (Refer to Axion in previous topic) |
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| = x.x + x.x' Axiom 5(b) |
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| = x. (x + x') Axiom 4(a) |
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| = x.1 Axiom 5(a) |
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| = x Axiom 2(b) |
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| Theorem 2(a) (Boundedness Law): |
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| x + 1 = 1 |
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| x + 1 = 1.(x + 1) 2(b) (Refer to Axion in previous topic) |
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| = (x +x') . (x + 1) Axiom 5(a) |
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| = x + x'.1 Axiom 4(b) |
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| = x + x' Axiom 2(b) |
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| = 1 Axiom 5(a) |
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| Theorem 2(b) : |
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| x.0 = 0 |
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| x.0 |
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| = x.0 + 0 2(a) (Refer to Axion in previous topic) |
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| = x.0 + x.x' Axiom (5) |
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| =x(0 + x') Axiom 4(a) |
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| = x.x' Axiom 2(a) |
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| = 0 Axiom (5) |
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| Theorem 3 (Involution Law): |
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| (x')' = x |
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| We know that x' is the complement of x. |
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| If x + x' = 1 and x.x' = 0, then |
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x + x' = 1 x' + x =1 and x. x' = 0 x'.x = 0 (3a and 3b) |
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x is the complement of x' |
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(x')' = x |
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| Theorem 4(a) (Absorption Laws): |
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| x + x.y = x |
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| x + x.y = x.1 + x.y by 2(b) (Refer to Axion in previous topic) |
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| = x.(1 + y) Axiom 4(a) |
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| = x.(y + 1) Axiom 3(a) |
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| = x.1 Axiom 2(a) |
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| = x Axiom 2(b) |
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| Theorem 4(b): |
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| x.(x + y) = x |
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| x.(x+y) |
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| = (x + 0)(x + y) 2(a) (Refer to Axion in previous topic) |
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| = x + (0.y) Axiom 4(a) |
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| = x + (y.0) Axiom 3(b) |
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| = x + 0 (Theorem 2) |
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| = x Axiom 2(a) |
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| Theorem 5(a): |
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| If B is a Boolean algebra, for any x, y in B, then 0' = 1. |
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| x + 1 = 1 (Theorem 2a) |
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| In particular for x = 0, we have |
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| 0 + 1 = 1 …(1) |
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| x.0 = 0 (Theorem 2b) |
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| In particular, for x =1 |
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| 1.0 = 0 |
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0.1 = 0 (3a) …(2) |
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| From (1) and (2), we have |
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For 0 B, 0 + 1 = 1 |
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| 0.1 = 0 |
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1 is the complement of 0. |
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0' = 1 (5) |
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| Theorem 5(b): |
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| In a Boolean algebra B, we have |
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| 1' = 0 |
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| x + 1 = 1 (Theorem 2a) |
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| In particular, for x = 0 |
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| 0 + 1 =1 |
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1 + 0 = 1 …(1) (Axiom 3a) |
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| x . 0 = 0 |
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In particular for 1 B, we have |
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| 1 . 0 = 0 …(2) |
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| From (1) and (2), we have |
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| 0 is the complement of 1. (Axiom 5) |
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