1. 1 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 Banda.b is in B –Commutativity:a+b = b+aanda.b = b.a –Associativity:a+(b+c)=(a+b)+c anda.(b.c) = (a.b).c –Identitya+0 = aanda.1=a –Distributivity:a+(b.c) = (a+b).(a+c) and a.(b+c)=(a.b)+(a.c) –Complementarity: a+a’=1anda.a’=0 An algebraic structure

2. 2 Besides being an algebraic structure other useful axioms and theorems are –Null:x+1 = 1andx.0 = 0 –Idempotency:x+x = xandx.x = x –Involution:(x’)’ = x –Uniting:x.y+x.y’=xand(x+y).(x+y’)=x –Absorption:x+x.y=xandx.(x+y)=x –(another form):(x+y’).y=x.yand(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,.,+) Boolean algebra and its axioms and theorems

3. 3 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: f D (x1,x2,…,xn,0,1,+,.) = f(x1,x2,…,xn,1,0,.,+) –multiplication and factoring: (x+y).(x’+z) = x.z+x’.yandx.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) Duality axioms and theorems

4. 4 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’ Proving theorems

5. 5 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 And now try some harder problem

6. 6 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 Multiple forms and equivalence