@ jatin saini
if you take 2 pair then only one is possible so you are taking first implicants then essential right??

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Procedure for findin prime implicants is
1.find all possible iimplicants
2.then from those implicants which cover at least one term covered by that implicant only are essential prime implicants

Let there are 12 minterms in a function in which 8 minterms are covered by 2 Essential Prime Implicants. Each of the remaining 4 minterms have 2 Non- Essential Prime Implicants. Then the total number of minimal expressions is Answer is 16. Can anyone provide the solution to this problem.

Find the number of Essential prime implicants present in the K Map of the function f=Σ(2,3,5,7,8,12,13).Here the answer is 2,can anybody explain why it is 2?

Consider the Boolean function, F(w, x, y, z) = wy + xy + w̅xyz + w̅ x̅ y + xz + x̅y̅z̅. Which one of the following is the complete set of essential prime implicants? (A) w,y,xz,x̅z̅ (B) w,y,xz (C) y,x̅y̅z̅ (D) y,xz,x̅z After constructing the KMAP by finding out minterms, the circled terms contribute to Essential prime implicants, but i dont' see any such options, the Answer is given D