$ [D′ + AB′ + A′C + AC′D + A′C′D]′$
making it canonical SOP
D′ it has missing ABC and they can be in 8 possible combinations with D' = 0 at end
xxxD’
xxx0
0000 0
0010 2
0100 4
0110 6
1000 8
1010 10
1100 12
1110 14
AB′ has missing CD and it will have four combination with
AB′xx
10xx
1000 8
1001 9
1010 10
1011 11
A′C has missing BD and it will have four combination with
A′xCx
0x1x
0010 2
0011 3
0110 6
0111 7
AC′D has missing B and it will have twocombination with
AxC′D
1x01
0001 1
0101 5
A′C′D has missing B and it will have twocombination with
A′xC′D
0x01
0001 1
0101 5
$=Σ(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14)’$
$= Σ(15) =ABCD $
i.e. only 1 term