AP CS De Morgan's Rules
!!A is equivalent to A !((p && q) || (q || r)) = ((!p || !q) && (!q && !r)) !((!p && q) || (q || !r)) = ((p || !q) && (!q && r)) is equivalent to:
!!A
A
!((p && q) || (q || r)) = ((!p || !q) && (!q && !r)) !((!p && q) || (q || !r)) = ((p || !q) && (!q && r))
is equivalent to: