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Boolean Algebra

Boolean Algebra. Supplementary material. A Boolean algebra. B is a boolean algebra means B is a set with elements in it a,b,…,0,1Є B and operators * (and), + (or) and ‘ (not) following the axioms Closure: if a,b Є B then a+b Є B , a*b Є B and a’ Є B

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Boolean Algebra

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  1. Boolean Algebra Supplementary material

  2. A Boolean algebra • B is a boolean algebra means B is a set with elements in it a,b,…,0,1Є B and operators * (and), + (or) and ‘ (not) following the axioms • Closure: if a,b Є B then a+b Є B , a*b Є B and a’ Є B • Commutative: if a,b Є B then a+b=b+a and a*b=b*a • Distributive: if a,b,c Є B thena*(b+c)=a*b+a*c and a+(b*c) =(a+b)*(a+c) • Identitity elements: there are elements 0,1 Є B such that 1*a=a and 0+a=a • Complementary Law: for aЄ B a+a’=1 and a*a’=0

  3. Theorems (to prove) • Associative law. For a,b,c Є B a*(b*c)=(a*b)*c and a+(b+c)=(a+b)+c • Idempotent property: for a Є B, a+a=a and a*a=a • Identity elements act as null elements: For a Є B, a+1=1 and a*0=0 • Involution: For a Є B, (a’)’ =a • Absorption: for a,b Є B, a+(a*b)=a and a*(a+b)=a • DeMorgan’s Laws: for a,b Є B, (a*b)’=a’+b’ and (a+b)’=a’*b’

  4. Some sample proofs • a+a=a • Proof: • a=0+a • =a*a’+a • =a+a*a’ • (a+a)*(a+a’) • (a+a)*1 • a+a

  5. More sample proofs • a+1=1 • Proof: • 1=a+a’ • =a+1*a’ • =(a+1)*(a+a’) • =(a+1)*1 • 1*(a+1) • a+1

  6. Sample proofs continued • a*0=0 • Proof • a*0=0+a*0 • a*a’+a*0 • a*(a’+0) • a*(0+a’) • a*a’ • 0

  7. Absorption: proof of one of them • a+a*b=a • Proof • a+a*b=1*a+a*b • =a*1+a*b • =a*(1+b) • =a*1 • =1*a • =a

  8. DeMorgan • (a*b)’=a’+b’ • proof: • This is a complicated proof that requires two parts. • Notice that the theorem seems to say that if you negate the parenthesized expression on the LHS you wind up with the RHS, sort of like x’=z, (which is something like saying the negation of x is z). We might use this idea. First, x and z should behave like “opposites”, that is, if we take x and + it with z we should get 1 and x*z should be 0. • Also, we’ll need to show that if two booleans behave this way, then they are “opposites”, that is, if x+z=1 and x*z=0 then x’=z

  9. DeMorgan continued • Show if x+z=1 and x*z=0 then x’=z. • (proof) • x’ =1*x’ • =(x+z)*x’ • =x’*(x+z) • =x’*x+x’*z • =0+x’*z • x*z+x’*z • z*(x+x’) • z

  10. Completeness • A set of operators is complete iff all switching functions of n variables can be written using only that set. • The set {*,+,’} is complete. • DNF==SOP and CNF==POS are canonical representations of switching functions. They use the idea of Duals – one is the dual of the other.

  11. Generating switching functions using DNF • Start with the table for f • For each row in f which has a value of 1, generate one term in the DNF and put an ‘or’ symbol (+) between each term. • Generate the terms as follows, if the variable v is a 0 in this row, put v’ into the product, if v is a 1 in this row, put v in the product.

  12. Example f(a,b,c)

  13. DNF (or POS) for f • f’s DNF has 4 terms, corresponding to the numerical values – rows - 0, 4, 5, 6 • f=a’*b’*c’+a*b’*c’+a*b’*c+a*b*c’

  14. CNF • To get the cnf use the dual concept. Generate a term (a sum, though) for each 0-valued row of f. For variable v a 0 add v to the sum, if v is a 1 add v’ to the sum. Conjunct all terms.

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