An algebraic structure consists of - PowerPoint PPT Presentation

1 / 6
About This Presentation
Title:

An algebraic structure consists of

Description:

Duality. A dual of a Boolean expression is derived by replacing. ... (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 ... – PowerPoint PPT presentation

Number of Views:38
Avg rating:3.0/5.0
Slides: 7
Provided by: arunks
Category:

less

Transcript and Presenter's Notes

Title: An algebraic structure consists of


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 ab is in B and a.b is in B
  • Commutativity ab ba and a.b b.a
  • Associativity a(bc)(ab)c and a.(b.c)
    (a.b).c
  • Identity a0 a and a.1a
  • Distributivity a(b.c) (ab).(ac) and
    a.(bc)(a.b)(a.c)
  • Complementarity aa1 and a.a0

2
Boolean algebra and its axioms and theorems
  • Besides being an algebraic structure other useful
    axioms and theorems are
  • Null x1 1 and x.0 0
  • Idempotency xx x and x.x x
  • Involution (x) x
  • Uniting x.yx.yx and (xy).(xy)x
  • Absorption xx.yx and x.(xy)x
  • (another form) (xy).yx.y and (x.y)yxy
  • de Morgans (xy..)x.y... and
    (x.y)xy.
  • Generalized de Morgans
  • 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 (xy)Dx.y and (x.y.)D xy
  • general duality fD(x1,x2,,xn,0,1,,.)
    f(x1,x2,,xn,1,0,.,)
  • multiplication and factoring
  • (xy).(xz) x.zx.y and x.yx.z(xz).(xy)
  • Consensus
  • (x.y)(y.z)(x.z)x.yxz and (xy).(yz).(xz)
    (xy).(xz)

4
Proving theorems
  • Prove x.yx.yx
  • distributivity x.yx.y x.(yy)
  • complementarity x.(yy) x.(1)
  • identity x.(1) x
  • Prove xx.y x
  • identity xx.y x.1x.y
  • distributivity x.1x.y x.(1y)
  • identity x.(1y) x.(1)
  • identity x.(1) x
  • NOR is equivalent of AND
  • (xy) x.y
  • NAND is equivalent of OR
  • (x.y) xy

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 (xxx)
  • 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 (aa).b.cin (bb).a.cin
    a.b.(cincin)
  • 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
Write a Comment
User Comments (0)
About PowerShow.com