The second prime implicant can be 'covered' by the third and fourth, and the third prime implicant can be 'covered' by the second and first, and neither is thus essential. If a prime implicant is essential then, as would be expected, it is necessary to include it in the minimized boolean equation. In some cases, the essential prime implicants do not cover all minterms, in which case additional procedures for chart reduction can be employed. The simplest "additional procedure" is trial and error, but a more systematic way is Petrick's method. In the current example, the essential prime implicants do not handle all of the minterms, so, in this case, the essential implicants can be combined with one of the two non-essential ones to yield one equation: