Simplify the Boolean function F = A′B′C′D′ + A′B′CD′ + A′BC′D + A′BCD′ + AB′C′D′ + AB′C′D + ABC′D′ + ABCD to a minimal sum-of-products form.

Step-by-step solution

1. 1

   Convert the expression to a minterm list. Reading each product as a $4$-bit pattern $ABCD$ (a primed literal is $0$):

   $A'B'C'D' = m_0, \ A'B'CD' = m_2, \ A'BC'D = m_5, \ A'BCD' = m_6$

   $AB'C'D' = m_8, \ AB'C'D = m_9, \ ABC'D' = m_{12}, \ ABCD = m_{15}$

   so

   $F = \Sigma m(0, 2, 5, 6, 8, 9, 12, 15)$

   The starting cost is $8$ terms and $32$ literals.

2. 2

   Enumerate the prime implicants. Combining adjacent minterms (those differing in exactly one bit) and keeping only the maximal groups gives seven prime implicants:

   $A'B'D'\,(0,2), \quad A'CD'\,(2,6), \quad B'C'D'\,(0,8), \quad AC'D'\,(8,12)$

   $AB'C'\,(8,9), \quad A'BC'D\,(5), \quad ABCD\,(15)$

   Every group has only two cells — there is no $4$-cell block anywhere, because no set of four of these minterms forms a K-map square.

3. 3

   Identify the essential terms. Minterm $m_5$ is covered only by $A'BC'D$ and $m_{15}$ only by $ABCD$, so those two full minterms are essential and must appear. Similarly $m_6$ appears only in $A'CD'$, $m_9$ only in $AB'C'$, and $m_{12}$ only in $AC'D'$ — three more essential prime implicants.

4. 4

   Cover what is left. The five essential terms $A'BC'D$, $ABCD$, $A'CD'$, $AB'C'$, $AC'D'$ together cover $\{2,5,6,8,9,12,15\}$, leaving only $m_0$. It can be picked up by either $A'B'D'$ or $B'C'D'$, both of size two, so there are exactly two minimal covers of equal cost.

5. 5

   Write the minimal SOP. Choosing $A'B'D'$:

   $F = A'B'D' + A'CD' + A'BC'D + AB'C' + AC'D' + ABCD$

   The equally minimal alternative replaces $A'B'D'$ with $B'C'D'$. Either way the cost is $6$ terms and $20$ literals, down from $8$ terms and $32$ literals.

6. 6

   Verify against the full truth table. Evaluating the minimised expression at all $16$ input combinations reproduces $1$ exactly on $\{0,2,5,6,8,9,12,15\}$ and $0$ elsewhere. The reason the reduction is modest is structural: minterms such as $m_5$ ($0101$) and $m_{15}$ ($1111$) have no neighbour in the set, so they survive as full four-literal products no matter how the map is grouped.