Question

Rewrite the following statement so that negations appear only within predicates (i.e., no negation or not is outside a quantifier or a compound statement using logical operators/connectives). Please write it clearly and show every step!

V ((Vy32 -P(x, y, z)) 4 (3 z Vy R(x, y, z)))

Answer 1

uopezəu əfqnoa ((((z**x)ukaze)^((z**x)d–zExa)) (((z*d*x) z—1ęza)^((z**x)dzA^E)))«E= szəriguend 10j mej s ued10W Əd ((((2*x+x\{A-E)^((z*4*x)d–zExa))(((z*4*x)7–1ęza)^((2*4*x)d_2AE)))*E = uoqjesə əfqnod ((((z*4+x)=ía-E)^((z**x)d=zE4A))+(((2*4*x)}(AZE)«^((2*+x)d=zExa))) E= De Morgan's law ==x[(-(Vy3z-P(x,y,z)) v-(=zVyR(x,y,z)))^(--(Vy2z=P(x,y,z))*--(Z2VyR(x,y,z)))) De Morgan's law ((( z*d*x)y<A-E)— v((2*1*x)d–zera)-)— v(((z*a*r)u«Aze) v((z** *x)d–zĘ(A)))«E= szəyğuend joj mej s ueblow əd ((((z**x)u«a-E)—v ((z**x)d–)-)^(((2*x*x)d«AE) ((z**x)d–zęđa ) –= меј әриәгелinbg ((( z*a*x)x+A+E)— v((z**x)d=zE1A)-)^(((2*x+x)\A-E) (*~*x)d–zE~a)))?A== (((2*x*x)d«AzE) - ((2*w*x)d=zE1A)) xa–