C959 WGU unit 3 questions and answers.

boolean complement denoted with bar symbol reverses elements value (bar over)0=1 (bar over)1=0 boolean addition 0+0=0 0+1=1 1+0=1 1+1=1 boolean multiplication 0×0=0 0×1=0 1×0=0 1×1=1 ⊕ exclusive or XOR outputs 1 only when inputs are different 0⊕0=0 1⊕0=1 0⊕1=1 1⊕1=0 idempotent of boolean x+x=? x*x=? x+x=x x*x=x associative of boolean (x+y)+z=? (xy)z=? (x+y)+z=x+(y+z) (xy)z=x(yz) commutative of boolean x+y=? xy=? x+y=y+x xy=yx distributive of boolean x+yz=? x(y+z)=? x+yz=(x+y)+(x+z) x(y+z)=xy+xz identity of boolean x+0=? x*1=? x+0=x x*1=x domination of boolean x*0=? x+1=? x*0=0 x+1=1 complement of boolean x(bar over)x=? x+(bar over)x=? x(bar over)x=0 (bar over)1=0 x+(bar over)x=1 (bar over)0=1 de morgan's of boolean (bar over)(x+y)=? (bar over)(xy)=? (bar over)(x+y)=(bar over)x*(bar over)y (bar over)(xy)=(bar over)x+(bar over)y absorption of boolean x+(xy)=? x(x+y)=? x+(xy)=x x(x+y)=x minterm literals that evaluate to 1 DNF disjunctive normal form expression that's the sum of products of literal c1+c2+...cm CNF conjunctive normal form expression that's product of sums of literals d1*d2*...dm complement form for DNF and CNF can only be applied to single variable multiplication in DNF can only be applied to single terms (one literal) addition in CNF can only apply to single literals (one term) NAND "not and" (bar over)xy = x↑y 0↑0=1 0↑1=1 1↑0=1 1↑1=0 NOR "not or" (bar over)(x+y) = x↓y 0↓0=1 0↓1=0 1↓0=0 1↓1=0 SAT boolean satisfiability takes boolean expression as input and asks if it's possible to set values of variables so that expression = 1 OR gate curved D boolean addition AND gate D boolean multiplication inverter gate triangle with circle computes complement NAND gate D with circle x↑y outputs 0 if all inputs = 1 outputs 1 if one or all input = 0