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An algebraic structure

An algebraic structure. An algebraic structure consists of a set of elements B binary operators {+, .} and a unary operator { ‘ } Such that following holds Membership: B contains at least two elements a and b Closure: a+b is in B and a.b is in B Commutativity: a+b = b+a and a.b = b.a

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An algebraic structure

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  1. An algebraic structure • An algebraic structure consists of • a set of elements B • binary operators {+, .} • and a unary operator { ‘ } • Such that following holds • Membership: B contains at least two elements a and b • Closure: a+b is in B and a.b is in B • Commutativity: a+b = b+a and a.b = b.a • Associativity: a+(b+c)=(a+b)+c and a.(b.c) = (a.b).c • Identity a+0 = a and a.1=a • Distributivity: a+(b.c) = (a+b).(a+c) and a.(b+c)=(a.b)+(a.c) • Complementarity: a+a’=1 and a.a’=0

  2. Boolean algebra and its axioms and theorems • Besides being an algebraic structure other useful axioms and theorems are • Null: x+1 = 1 and x.0 = 0 • Idempotency: x+x = x and x.x = x • Involution: (x’)’ = x • Uniting: x.y+x.y’=x and (x+y).(x+y’)=x • Absorption: x+x.y=x and x.(x+y)=x • (another form): (x+y’).y=x.y and (x.y’)+y=x+y • de Morgan’s: (x+y+..)’=x’.y’... and (x.y…)’=x’+y’+…. • Generalized de Morgan’s: • f’(x1,x2,…,xn,0,1,+,.) = f(x1’,x2’,…,xn’,1,0,.,+)

  3. Duality axioms and theorems • Duality • A dual of a Boolean expression is derived by replacing . by +, + by ., 0 by 1, and 1 by 0 and leaving variables unchanged • Any theorem that can be proven is thus also proven for • a meta-theorem (a theorem about theorems) • duality: (x+y+…)D=x.y…… and (x.y….)D = x+y+… • general duality: fD(x1,x2,…,xn,0,1,+,.) = f(x1,x2,…,xn,1,0,.,+) • multiplication and factoring: • (x+y).(x’+z) = x.z+x’.y and x.y+x’.z=(x+z).(x’+y) • Consensus: • (x.y)+(y.z)+(x’.z)=x.y+x’z and (x+y).(y+z).(x’+z)=(x+y).(x’+z)

  4. Proving theorems • Prove x.y+x.y’=x • distributivity: x.y+x.y’ = x.(y+y’) • complementarity: x.(y+y’) = x.(1) • identity: x.(1) = x • Prove x+x.y = x • identity: x+x.y = x.1+x.y • distributivity: x.1+x.y = x.(1+y) • identity: x.(1+y) = x.(1) • identity: x.(1) = x • NOR is equivalent of AND • (x+y)’ = x’.y’ • NAND is equivalent of OR • (x.y)’ = x’+y’

  5. And now try some harder problem • Simplify Boolean expression for carry function in a 3-bit adder • Cout = a’.b.cin + a.b’.cin + a.b.cin’ + a.b.cin • Each of the first, second, and third term can be combined with the last term • Use identity to make copies of the last term 3 times (x+x=x) • Cout = a’.b.cin + a.b’.cin + a.b.cin’ + a.b.cin + a.b.cin + a.b.cin • Use associativity to bring terms together • Cout = a’.b.cin + a.b.cin + a.b’.cin + a.b.cin + a.b.cin’ + a.b.cin • Then use distributivity to combine terms • Cout = (a’+a).b.cin + (b’+b).a.cin + a.b.(cin’+cin) • Next use complementarity to reduce • Cout = (1).b.cin + (1).a.cin + a.b.(1) • Finally using identity gives Cout = b.cin + a.cin + a.b

  6. Multiple forms and equivalence • Canonical Sum-of-Product form • Canonical Product-of-sum form • How to convert one from other? • Minterm expansion of F to minterm expansion of F’ • Just take the terms that are missing • Maxterm expansion of F to maxterm expansion of F’ • Just take the terms that are missing

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