Boolean Algebra

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,,0,1? B and operators
  (and), (or) and (not) following the axioms
• Closure if a,b ? B then ab ? B , ab ? B and
  a ? B
• Commutative if a,b ? B then abba and abba
• Distributive if a,b,c ? B then a(bc)abac
  and a(bc) (ab)(ac)
• Identitity elements there are elements 0,1 ? B
  such that 1aa and 0aa
• Complementary Law for a? B aa1 and aa0

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Theorems (to prove)

• Associative law. For a,b,c ? B a(bc)(ab)c
  and a(bc)(ab)c
• Idempotent property for a ? B, aaa and aaa
• Identity elements act as null elements For a ?
  B, a11 and a00
• Involution For a ? B, (a) a
• Absorption for a,b ? B, a(ab)a and a(ab)a
• DeMorgans Laws for a,b ? B, (ab)ab and
  (ab)ab

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Some sample proofs

• aaa
• Proof
• a0a
• aaa
• aaa
• (aa)(aa)
• (aa)1
• aa

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More sample proofs

• a11
• Proof
• 1aa
• a1a
• (a1)(aa)
• (a1)1
• 1(a1)
• a1

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Sample proofs continued

• a00
• Proof
• a00a0
• aaa0
• a(a0)
• a(0a)
• aa
• 0

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Absorption proof of one of them

• aaba
• Proof
• aab1aab
• a1ab
• a(1b)
• a1
• 1a
• a

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DeMorgan

• (ab)ab
• 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 xz,
  (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 xz
  should be 0.
• Also, well need to show that if two booleans
  behave this way, then they are opposites, that
  is, if xz1 and xz0 then xz

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DeMorgan continued

• Show if xz1 and xz0 then xz.
• (proof)
• x 1x
• (xz)x
• x(xz)
• xxxz
• 0xz
• xzxz
• z(xx)
• z

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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.
• DNFSOP and CNFPOS are canonical
  representations of switching functions. They use
  the idea of Duals one is the dual of the other.

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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.

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Example f(a,b,c)
a b c f
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0
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DNF (or POS) for f

• fs DNF has 4 terms, corresponding to the
  numerical values rows - 0, 4, 5, 6
• fabcabcabcabc

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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.