Use minimum number of half adders for the boolean expression
Let's simplify the expression with a brute force:


As we see the result of the function almost always equal to A except two cases:

C=1 and D=1 (the result is 1)
B=1 and C=1 and D=0 (the result is 0)

So we can write a new expression F=(A+CD)(BCD')'.
Now count half-adders:

CD - 1 half-adder
A+CD - 2 half-adders (OR = two half adders)
BC - 1 half-adder
D' - 1 half-adder (NOT = one half adder, one input is tie to 1)
BC and D' - 1 half-adder
(BCD')' - 1 half-adder
(A+CD)(BCD')'- 1 half-adder

Total: 8 half-adders