Answer to Question #158972 in C++ for Banele

Question

prove the following using the laws algebra of propositions

(a) ~[x or (~x and y)] =~x and ~y

Answer 1

~[x or (~x and y)] =(de Morgan's law)= ~x and ~(~x and y) =(de Morgan's law)= ~x and (~~x or ~y)

=(double negation)= ~x and (x or ~y) =(distributivity of and over or)= (~x and x) or (~x and ~y)

=(indempotence of and)= true or (~x and ~y) =(annihilation for or)= ~x and ~y