Given that ,Tk,n(x) =1 if ∑ xi ≥ k ,which is eqt to say ,Tk,n(x) =1 if atleast k of its n inputs are 1.
We need to have atleast k inputs whose values are 1 for Tk,n(x) =1.
y ∈ {y1,y2,y3,.........y(i-1),y(i+1),........yn}
In this yi is misssing and we need to findout the value of Tk,n−1(y1,y2,....yi−1,yi+1,...,yn).
See, we dont know what is the value of yi but we know that it is either 0 or 1 because codomain is {0,1}.
And in Qs given that y has exactly k - one's and (n-k) - zero's.
Missing yi has only two possible value either 0 or 1.
CASE-1: if yi = 0
It means k inputs are present in (y1,y2,....yi−1,yi+1,...,yn) whose values are 1.It satisfy the condition Tk,n(x) =1 if atleast k of its n inputs are 1.
Therefore, Tk,n−1(y1,y2,....yi−1,yi+1,...,yn) = 1 = complement of yi
CASE-2: if yi = 1
It means only (k-1) inputs are present in (y1,y2,....yi−1,yi+1,...,yn) whose values are 1.So it does not satisfy the condition Tk,n(x) =1 if atleast k of its n inputs are 1.
Therefore, Tk,n−1(y1,y2,....yi−1,yi+1,...,yn) = 0 = complement of yi
Hence,it is equivalent to say Tk,n−1(y1,y2,....yi−1,yi+1,...,yn) = complement of yi .
The correct answer is,(d) ¬yi