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11. Boolean Algebra GapFill

Target Level
C
Running Total
0
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Attempt
1 of 3

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Simplifying Boolean expressions is usually done in stages, and there are several sets of rules that can be followed at each stage of simplification.   Double negationDistributionSubstitutionIdentity essentially means that where there are two consecutive  ANDXORORNOT operators, they can both be disregarded; for example,  ¬¬A = ¬A¬A = ¬¬A¬¬A = A¬A = A.  Then there's  distributionidentitysubstitutionassociation, which is typically used for the removal of brackets, often in order to facilitate subsequent simplification.  An example of this rule in application is
A∧(B∨C) =  A∨B∨A∨CA∨B∧A∨CA∧B∨A∧CA∧B∧A∧C  Substitution lawsDe Morgan's lawsCommutative lawsAssociative laws allow a NOT operator, applied to a compound expression, to be applied to individual elements of that expression; for example, ¬(A∧B∧C) =  ¬A∧¬B∧¬C¬A∧B∧C¬A∨B∨C¬A∨¬B∨¬C.

 AssociativeDouble negationDistributionCommutative law deals with the order in which values or propositional letters appear around an operator.  An example of this would be  A∧B∧C = C∨B∨AA∧B∧C = C∧B∧AA∨B∧C = C∧B∧AA∧B∧C = C∧B∨A or  ¬A∨B = ¬B∨AA∨B = B∨AA∧B = B∨AA∨B = B∧A  DistributionIdentityDe Morgan'sAssociative law is similar, in that it reorders rather than simplifies, except this law deals with the positioning of brackets.  In mathematics, you should already be familiar with the fact that  A-(B*C) = (A-B)*CA*(B*C) = (A*B)*CA+(B*C) = (A+B)*CA+(B/C) = (A+B)/C; in Boolean algebra, according to this law,  D∧(E∧F) = (D∧E)∧FD∧(¬E∧F) = (¬D∧E)∧FD∧(E∨F) = (D∧E)∨FD∨(E∧F) = (D∨E)∧F.

This is your 1st attempt! You get 3 marks for each one you get right. Good luck!

Pass Mark
72%